292 Where e is the induced emf and di dt is the rate of change of current. To maintain the flow of current in the circuit, applied emf (e) must be equal and opposite to the induced emf (e'). According to Kirchhoff's voltage law as there is no resistance in the circuit, e = - e' fi ff ffl fflffi fl ff fi ff e L di dt L di dt di e L dt (from Eq. (13.6)) Integrating the above equation on both the sides, we get, di e L dt i e t L dt e e t i e L t fi fi fi ff ff ff ff ffiffl fl 0 0 0 sin ( sin ) cos constant Constant of integration is time independent and has the dimensions of i. As the emf oscillates about zero, i also oscillates about zero so that there cannot be any component of current which is time independent. Thus, the integration constant is zero fi ff ffl ffl ffi fl ffl ffi fl ff ffi fl i e L t t t 0 2 2 sin sin cos fi ff ffl ff ffl i e L t i i t 0 0 2 2 sin sin --- (13.7) where i e L 0 0 fi ff --- (13.8) where i 0 is the peak value of current. Eq. (13.7) gives the alternating current developed in a purely inductive circuit when connected to a source of alternating emf. Comparing Eq. (13.5) and (13.7) we find that the alternating current i lags behind the alternating voltage emf e by a phase angle of π/2 radians (90°) or the voltage across L leads the current by a phase angle of π/2 radians (90°) as shown in Fig. 13.7. Fig 13.7: Graph of e and i versus ωt. Fig. 13.8 Phasor diagram for purely inductive. Phasor diagram: The phasor representing peak emf e0 makes an angle ωt in anticlockwise direction from horizontal axis. As current lags behind the voltage by 90°, so the phasor representing i 0 is turned clockwise with the direction of e0 as shown in Fig. 13.8. Inductive Reactance (XL): The opposing nature of inductor to the flow of alternating current is called inductive reactance. Comparing Eq (13.8) with Ohm's law, i e R 0 0 = we find that ω L represents the effective resistance offered by the inductance L, it is called the inductive reactance and denoted by XL. ∴ XL = ω L = 2πfL. (. . .ω = 2π/T = 2πf) where f is the frequency of the AC supply. The function of the inductive reactance is similar to that of the resistance in a purely resistive circuit. It is directly proportional to the inductance (L) and the frequency (f) of the alternating current.
293 The dimensions of inductive reactance is the same as those of resistance and its SI unit is ohm (Ω). In DC circuits f = 0 ∴XL = 0 It implies that a pure inductor offers zero resistance to DC, i.e., it cannot reduce DC. Thus, its passes DC and blocks AC of very high frequency. In an inductive circuit, the self induced emf opposes the growth as well as decay of current. The current flowing in the circuit transfers charge to the plates of the capacitor which produces a potential difference between the plates. As the current reverses its direction in each half cycle, the capacitor is alternately charged and discharged. Suppose q is the charge on the capacitor at any given instant t. The potential difference across the plates of the capacitor is V q C= or q = CV --- (13.10) The instantaneous value of current (i) in the circuit is i dq dt d dt = = ( ) Ce (. . . V = e at every instant) fi fi ff ffl fi ffi fi fl fi d dt e t e e t e t e C t i e ( sin ) sin / C C 0 0 0 0 0 1 cos cos 1 2 2 / sin cos sin C t t t fi --- (13.11) The current will be maximum when sin (ωt + π/2) = 1, so that i = i 0 where, peak value of current is i e C 0 0 1 fi /ff --- (13.12) fi ff ffl ffi fl i i t 0 2 sin --- (13.13) Fig: 13.9 An AC source connected to a capacitor. Example 13.3: An inductor of inductance 200 mH is connected to an AC source of peak emf 210 V and frequency 50 Hz. Calculate the peak current. What is the instantaneous voltage of the source when the current is at its peak value? Solution: Given L = 200 mH =0.2 H e0 = 210 V f = 50 Hz Peak Current i 0 = e XL 0 = e fL 0 2π = . . 210 2 3 × × 142 50× 0 2 ∴ i 0 = 3.342 A As in an inductive AC circuit, current lags behind the emf by π 2 , so the voltage is zero when the current is at its peak value. c) AC voltage applied to a capacitor: Let us consider a capacitor with capacitance C connected to an AC source with an emf having instantaneous value e = e0 sin ω t --- (13.9) This is shown in Fig. 13.9 Fig. 13.10 Graph of e and i versus ωt. From Eq. (13.9) and Eq. (13.13) we find that in an AC circuit containing a capacitor only, the alternating current i leads the alternating emf e by phase angle of π/2 radian as shown in Fig. 13.10.
294 Phasor diagram: are the same as that of resistance and its SI unit is ohm (Ω). Table 13.1: Comparison between resistance and reactance. Resistance Reactance Equally effective for AC and DC Current is affected (reduced) but energy is not consumed (heat is not generated). The energy consumption by a coil is due to its resistive component. Its value is independent of frequency of the AC Inductive reactance (XL = 2π fL) is directly proportional and capacitive reactance Xc fC fi ff ffl ffi fl 1 2 is inversely proportional to the frequency of the AC. Current opposed by a resistor is in phase with the voltage. Current opposed by a pure inductor lags in phase while that opposed by a pure capacitor leads is phase by πc over the voltage. The phasor representing peak emf makes an angle ω t in anticlockwise direction with respect to horizontal axis. As current leads the voltage by 90°, the phasor representing i 0 current is turned 90° anticlockwise with respect to the phasor representing emf e0 . The projections of these phasors on the vertical axis gives instantaneous values of e and i. Capacitive Reactance: The instantaneous value of alternating current through a capacitor is given by i e C i fi ff ffl ffi fl fi ff ffl ffi fl 0 0 1 2 2 ( / ) t t sin sin Comparing Eq. (13.12) with Ohm's law, i e R 0 0 = we find that (1/ω C) represents effective resistance offered by the capacitor called the capacitive reactance denoted by XC. fi ff X ff C fC C 1 1 ffl ffi2 where f is the frequency of AC supply. The function of capacitive reactance in a purely capacitive circuit is to limit the amplitude of the current similar to the resistance in a purely resistive circuit. XC varies inversely as the frequency of AC and also as the capacitance of the condenser. In a DC circuit, f = 0 ∴XC = ∞ Thus, capacitor blocks DC and acts as open circuit while it passes AC of high frequency. The dimensions of capacitive reactance Fig.13.11: Phasor diagram for purely capacitive circuit. Example 13.4: 4. A Capacitor of 2 µF is connected to an AC source of emf e = 250 sin 100πt. Write an equation for instantaneous current through the circuit and give reading of AC ammeter connected in the circuit. Solution: Given C = 2µF = 2 fi ff 10 6 F e0 = 250 V ω = 100π rad/sec The instantaneous current through the circuit i = i 0 sin (ωt + π 2 ) = ωC e0 sin (ωt + π 2 ) = 3.142 × 2 × 10-4 × 250 sin (100πt + π 2 )
295 (d) AC circuit containing resistance inductance and capacitance in series (LCR circuit): Above we have studied the opposition offered by a resistor, pure inductor and capacitor to the flow of AC current independently. Now let us consider the total opposition offered by a resistor, pure inductor and capacitor connected in series with the alternating source of emf as shown in Fig. 13.12. As eR is in phase with current i 0 the vector eR is drawn in the same direction as that of i, along the positive direction of X-axis represented by OA fi ff fifi . The voltage across L and C have a phase different of 180° hence the net reactive voltage is (eL - eC). Assuming eL > eC represented by OB′ in the figure. The resultant of OA fi ff fifi and OB' fi ff fififi is the diagonal OK of the rectangle OAKB' fi ff ffl ff ffl fl ffi ff ffl fl ffi OK OA OB 2 2 0 2 2 2 0 0 2 e e e e iX iX R L C L C ( ) i R0 e R X X e R X X e Z L C L C 0 2 2 0 2 2 0 ff ffl fl ffi fi ff ffl ffi fl ff i i i 0 0 0 Comparing the above equation with the relation V i = R , the quantity Z R fi ff L C ffl 2 2 (X X ) represents the effective opposition offered by the inductor, capacitor and resistor connected in series to the flow of AC current. This total effective resistance of LCR circuit is called the impedance of the circuit and is represented by Z. The reciprocal of impedance of an AC circuit is called admittance. Its SI unit is ohm-1 or siemens. It can be defined as the ratio of rms voltage to the rms value of current Impedance is expressed in ohm (Ω). Phasor diagram: Fig. 13.12: Series LCR circuit. Let a pure resistor R, a pure inductance L and an ideal capacitor of capactance C be connected in series to a source of alternative emf. As R, L and C are in series, the current at any instant through the three elements has the same amplitude and phase. Let it be represented by i = i 0 sin ωt. The voltage across each element bears a different phase relationship with the current. The voltages eL, eC and eR are given by eR = iR, eL = iXL and eC = iXC As the voltage across the capacitor lags behind the alternating current by 90°, it is represented by OC fi ff fifi , rotated clockwise through 90° from the direction of i 0 . OC fi ff fifi is along OY′ in the phasor diagram shown in the phasor diagrams in Fig. 13.13. = 0.1571 sin (100πt + π 2 ) Reading of the AC ammeter is i rms= 0.707 i 0 = 0.707 × 0.1571 i ms = 0.111A Fig. 13.13: Phasor diagram for an LCR circuit.
296 From the phasor diagram (Fig. 13.13) it can be seen that in an AC circuit containing L, C and R, the voltage leads the current by a phase angle φ , tanfi ff ff ffl ff AK ffl OA = OB' OA e e e iX iX i R L C R o L o C o tan t fi fi ff an ffl ffi ff fl ffl X X ffl R X X R L C L C 1 ∴ The alternating current in LCR circuit would be represented by i = i 0 sin (ωt + φ ) and e = e0 sin (ωt + φ ) We can now discuss three cases based on the above discussion. (i) When XL = XC then tan φ = 0. Hence voltage and current are in phase. Thus the AC circuit is non inductive. (ii) When XL > XC, tan φ is positive ∴ φ is positive. Hence voltage leads the current by a phase angle φ The AC circuit is inductance dominated circuit. (iii)When XL < XC, tan φ is negative ∴ φ is negative. Hence voltage leads the current by a phase angle φ The AC circuit is capacitance dominated circuit. Impedance triangle: From the three phasors e i R L R e i XL e i C XC = = 0 0 , , = 0 we obtain the impedance triangle as shown in Fig 13.14. Fig. 13.14: Impedance triangle. Example 13.5: A 100mH inductor, a 25 μF capacitor and a 15 Ω resistor are connected in series to a 120 V, 50 Hz AC source. Calculate (i) impedance of the circuit at resonance (ii) current at resonance (iii) Resonant frequency Solution: Given L = 100 mH = 10-1H C = 25 μF = 25 x 10-6F R = 15Ω erms=120 V f = 50 Hz (i) At resonance, Z = R = 15Ω (ii) i rms= e R A rms = = 120 15 8 (iii) f = 1 2π LC = 1 2 3 142 10 25 10 1 6 . fi fi fi ff ff = 1 9 9356 10 3 . fi ff ∴ f = 100.65 Hz Example 13.6: A coil of 0.01H inductance and 1Ω resistance is connected to 200V, 50Hz AC supply. Find the impedance of the circuit and time lag between maximum alternating voltage and current. Solution: Given Inductance L = 0.01H Resistance R = 1Ω e0 = 200 V Frequency f = 50 Hz Impedance of the circuit Z = R XL 2 2 + = R fL 2 2 fi ff ffl 2ffi = 1 2 3 142 50 0 01 2 2 fi ff ffl ffi fi ff . . ffi ffi = 10 872. = 3.297Ω tan φ = ωL R = 2π fL R = 2 3 142 50 0 01 1 . × × × . = 3.142 φ = tanfi ff ffl 1 3 142. = 72.35° Phase difference, φ = 72 35 180 . fiff rad Time lag between maximum alternating voltage and current Δt = fi ff = 72 35 180 2 50 0 004 . . fi fi fi ff ffl ffl s The diagonal OK represents the impedance Z of the AC circuit. Z R fi ff ffl 2 2 ( ) X X L C , the base OA represents the Ohmic resistance R and the perpendicular AK represents reactance (XLXC). ∠AOK = φ , is the phase angle by which the voltage leads the current is the circuit, where tan X X L C fi ff ffl R
297 13.6 Power in AC circuit: We know that power is defined as the rate of doing work. For a DC circuit, power is measured as a product of voltage and current. But since in an AC circuit the values of current and voltage change at every instant the power in an AC circuit at a given instant is the product of instantaneous voltage and instantaneous current. a) Average power associated with resistance (power in AC circuit with resistance). In a pure resistor, the alternating current developed is in phase with the alternating voltage applied i.e. when e = e0 sin ωt then i = i 0 sin ωt Now instantaneous power P = ei. P = (e0 sin ωt) (i 0 sinωt ) = e0 i 0 sin2 ωt --- (13.14) The instantaneous power varies with time, hence we consider the average power for a complete cycle by integrating Eq. (13.14). fi ff Pav work done by the emf on the charges in one cycle time for one cycle ff ff ffl ffl Pdt T e i t dt T T T 0 0 0 2 0 sin ffi ff ff fl ffl e i T t dt e i T T t T 0 0 2 0 0 0 2 2 sin sin ffi ffi dt T e i P P e i T a rms rms ff ff fi ff ff ffl 2 2 2 0 0 0 v --- (13.15) P is also called as apparent power. b) Average power associated with an inductor: In an purely inductive circuit, the current lags behind the voltage by a phase angle of π/2. i.e., when e = e0 sin ω t then i = i 0 sin (ωt - π/2). Now, instantaneous power P = ei P = (e0 sin ωt) (i 0 sin (ωt - π/2)) = - e0 i 0 sin ωt cos ωt = - e0 i 0 sin ωt cos ωt fi ff ff ff ffl ffi P Pdt T e i a T v work done in one cycle time for one cycle 0 0 0 0 0 0 0 0 0 0 2 2 2 2 sin cos sin cos sin fl fl fl fl fl t t dt T e i t t dt T e i t T T T ffi ffi ff ff ffl ffi ff ffl ffl v ff dt e i T t P T a 0 0 0 2 2 2 0 cos fl fl ∴ average power over a complete cycle of AC through an ideal inductor is zero. c)Average power associated with a capacitor: In a purely capacitive circuit the current leads the emf by a phase angle of π/2 ie when e = e0 sin ωt then i = i 0 sin (ωt + π/2) i = i 0 cos ωt Now, instantaneous power P = ei = (e0 sin ωt) (i 0 cos ωt) = e0 i 0 sin ωt cos ωt. Example 13.7: A 100Ω resistor is connected to a 220V, 50Hz supply (i) What is the rms value current in the circuit? (ii) What is the net power consumed over a full cycle? Solution: Given R = 100Ω, erms= 220V , f = 50 Hz (i) i rms= e R A rms = = 220 100 2 2. (ii) Net Power Consumed Pav = erms.i rms = 220 × 2.2 = 484 W
298 t As seen above, =T / 2 dt = 0 fi ff ff sin cos sin 2 0 0 ffl ffl ffl t dt t t T T and --- (13.17) substituting (13.17) in (13.16) Pv = e i T T a 0 0 2 cos . fi fi sin ff ffl ffi fl = v ( ) cos . cos ( . 0 2 13 1 0 0 e i T T P e i a rms rms fi fi 8) This power (Pav) is also called as true power. The average power dissipated in the AC circuit of inductor. Capacitor and resistor connected in series not only depends on rms values of current and emf but also on the phase difference φ between them. The factor cosφ is called as power factor fi ff Power factor (cos true power ( ) apparent power av ffl) P P from impedance triangle ff ffi fl R R XL C X 2 2 fi ff Power factor cos ff resistance impedance ffl R Z In a non inductive circuit XL = XC fi ff Power factor (cos ffl) R R2 ff ff fi ff R R 1 0 ffl In a purely inductive and capacitive circuit; φ = 90° ∴ Power factor = 0 Average power consumed in a pure inductor or ideal capacitor Pav = erms i rms cos 90° = zero. ∴Current through pure inductor or ideal capacitor which consumes no power for its maintenance, in the circuit is called idle current or wattless current. Power dissipated in a circuit is due to resistance only. fi ff ff ff ffl P P Pdt T e a a T v v work done in one cycle time for one cycle 0 0 0 0 0 i t t T P T a sin c ffi ffi os ffl v ff afl s shown above Average power supplied to an ideal capacitor by the source over a complete cycle of AC is also zero. d) Average power in LCR Circuit: Let e = e0 sin ωt be the alternating emf applied across the series combination of pure inductor, capacitor and resistor as shown in Fig. 13.16. Fig. 13.15: LCR series circuit. There is a phase difference φ between the applied emf and current given by i = i 0 sin (ω t ± φ ) Instantaneous power is given by P = ei = (e0 sin ωt) i 0 sin (ω t ± φ ) = e0 i 0 [sin ωt cos φ ± cosωt sinφ ] sin ωt = e0 i 0 [sin2 ωt cos φ ± cos ωt sinφ sin ω ] ∴ Average power work done in one cycle time for one cycle Pv Pdt a fi fi 0 0 0 2 0 0 0 T T T e i t t t dt T e i T dt ff ff fi ffi ffl fl fi sin cos cos sin sin cos s in sin cos sin 2 0 0 t t t dt T T ff ff ffi fl ffl --- (13.16)
299 charged capacitor is connected to an inductor, the charge is transferred to the inductor and current starts flowing through the inductor. Because of the increasing current there will be a change in the magnetic flux of the inductor in the circuit. Hence induced emf is produced in the circuit. This self- induced emf will try to oppose the growth of the current. Due to this the charge (energy stored in) on the capacitor decreases and an equivalent amount of energy is stored in the inductor in the form of magnetic field. When the discharging of the capacitor completes, all the energy stored in the capacitor will be stored in the inductor. The capacitor will become fully discharged whereas inductor will be storing all the energy. As a result now the inductor will start charging the capacitor. The current and magnetic flux linked with the inductor starts decreasing. Therefore an induced emf is produced which recharges the capacitor in the opposite direction. This process of charging and discharging of capacitor is repeated and energy taken from the source keeps on oscillating between the capacitor (C) and the inductor (L). When a charged capacitor is allowed to discharge through a non-resistive inductor, electrical oscillations of constant amplitude and frequency are produced. These oscillations are called LC oscillations. This is explaind in Fig. 13. 16. Example 13.8: A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3Ω, L = 25.48 mH and C = 796 µf. Find i) The impedance of the circuit ii) The phase difference between the voltage across source and the currents iii) The power factor iv) The power dissipated in the surface Solution: Given e0 = 283 V, f = 50 Hz, R = 3Ω, L = 25.48 × 10-3 H, C = 796 × 10-6 F XL = 2πfL = 2 × 3.142 × 50 × 25.48 × 10-3 = 8Ω XC = 1 2π fL = 1 2 3 142 50 796 10 6 fi fi fi fi ff . = 1 0 2501. = 4Ω Therefore Z = R Xc c Y 2 2 fi ff ( ) = 3 8 4 2 2 fi ff ( ) = 5 Ω Phase difference φ is given by tan φ = X X R L c − = 8 4 3 − = 4 3 Therefore, φ = tan-1( 4 3 ) = 53.10 Thus the current lags behind the voltage across the source by a phase angle of 53.10 Power factor = cos φ = 0.6 Power dissipated in the circuit Pav = erms i rms cos φ = e0 2 e R 0 2 (0.6) = 283 2 283 2 3 × 0.6 = 8008.9 W 13.7 LC Oscillations: We have seen in chapters 8,10 and 12 that capacitors and inductors store energy in their electric and magnetic fields respectively. When a capacitor is supplied with an AC current it gets charged. When such a fully Fig. 13.16 (a) Let a capacitor with initial charge q0 at (t = 0) be connected to an ideal inductor (zero resistance). The electrical energy stored in the dielectric medium between the plates of the capacitor is UE = 1 2 q C 0 2 .Since there is no current in the circuit the energy stored in the magnetic field of the inductor is zero.
300 increases, it builds up a magnetic field around the inductor. A part of the electrical energy of the capacitor gets stored in the inductor in the form of magnetic energy U LI B = 1 2 2 entire energy is again stored as 1 2 0 2 q k in the electric field of the capacitor. The capacitor begins to discharge again sending current in opposite direction. The energy is once again transferred to the magnetic field of the inductor. Thus the process repeats itself in the opposite direction. The circuit eventually returns to the initial state. Thus the energy of the system continuously surges back and forth between the electric field of the capacitor and magnetic field of the inductor. This produces electrical oscillations of a definite frequency . These are called LC Oscillations. If there is no loss of energy the amplitude of the oscillations remain constant and the oscillations are undamped. However LC oscillations are usually damped due to following reasons. 1. Every inductor has some resistance. This causes energy loss as heat. The amplitude of oscillations goes on decreasing and they finally die out. 2. Even if the resistance were zero, total energy of the system would not remain constant. It is radiated away in the form of electromagnetic waves. Working of radio and TV transmitters is based on such radiations. 13.8 Electric Resonance: Have you ever wondered how radio picks certain frequencies so you can play your favourite channel or why does a glass break down in an orchestra concert? Why do you think you encounter such situations? The answer lies in the phenomenon of resonance. The phenomenon of resonance can be observed in systems that have a tendency to oscillate at a particular frequency, which is called the natural frequency of oscillation of the system. When such a system is driven by an energy source, whose frequency is equal to the natural frequency of the system, the amplitude Fig. 13.16 (b) As the circuit is closed, the capacitor begins to discharge itself through the inductor giving rise to a current (I) in the circuit. As the current (I) Fig. 13.16 (c) At a later instant the capacitor gets fully discharged and the potential difference across its plates becomes zero. The current reaches its maximum value I 0, the energy in the magnetic field is energy 1 2 0 2 LI . Thus the entire electrostatic energy of the capacitor has been converted into the magnetic field energy of the inductor. Fig. 13. 16 (d) After the discharge of the capacitor is complete, the magnetic flux linked with the inductor decreases inducing a current in the same direction (Lenz’s Law) as the earlier current. The current thus persists but with decreasing magnitude and charges the capacitor in the opposite direction. The magnetic energy of the inductor begins to change into the electrostatic energy of the capacitor. Fig. 13.16 (e) The process continues till the capacitor is fully charged with a polarity which is opposite to that in its initial state. Thus the
301 of oscillations become large and resonance is said to occur. (a) Series resonance circuit: At ω= ωr , value of peak current (i 0 ) is maximum. The maximum value of peak current is inversely proportional to R (fi fi i ) e R 0 0 = . For lower R values, i 0 is large and vice versa. The variation of rms current with frequency of AC is as shown in graph 13.18. The curve is called the series resonance curve. At resonance rms current becomes maximum. This circuit at resonant condition is very useful for radio and TV receivers for tuning the signal from a desired transmitting station or channel. Characteristics of series resonance circuit 1) Resonance occurs when XL = XC 2) Resonant frequency fr = 1 2π LC 3) Impedance is minimum and circuit is purely resistive. 4) Current has a maximum value. 5) When a number of frequencies are fed to it, it accepts only one frequency (fr ) and rejects the other frequencies. The current is maximum for this frequency. Hence it is called acceptor circuit. b) Parallel resonance circuit: A parallel resonance circuit consists of a coil of inductance L and a condenser of capacity C joined in parallel to a source of alternating emf. as shown in Fig. 13.19. A circuit in which inductance L, capacitance C and resistance R are connected in series (Fig. 13.17), and the circuit admits maximum current corresponding to a given frequency of AC, is called a series resonance circuit. The impedance (Z) of an LCR circuit is given by Z = R L C 2 2 1 fi ff ffl ffi fl At very low frequencies, inductive reactance XL= ωL is negligible but capacitive reactance XC= 1 ωC is very high. As we increase the applied frequency then XL increases and XC decreases. At some angular frequency (ωr ), XL = XC i.e. fi fi r r L C ff 1 ∴ ( ) ωr 2 = 1 LC or 2 2 fi f ff fflr ffi 1 LC ∴ 2πf r = 1 LC ∴ fr = 1 2 LC Where fr is called the resonant frequency. At this particular frequency f r , since XL = XC we get Z= R2 + 0 = R. This is the least value of Z Thus, when the impedance of on LCR circuit is minimum ,circuit is said to be purely resistive, current and voltage are in phase and hence the current i o = = e z e R 0 0 ff ff is maximum. This condition of the LCR circuit is called resonance condition and this frequency is called series resonant frequency. Fig. 13.18: Series resonance curve. Fig. 13.19 : Parallel resonance circuit. L Let the alternating emf supplied by the source be e = e0 sinω t Fig. 13.17: Series resonance circuit.
302 Do you know? In case of an inductor, the current lags behind the applied emf by a phase angle of π/2, then the instantaneous current through L is given by fi fi i / L fi ff ffl ffi e X t L 0 sin fl 2 Similarly in a capacitor ,as current leads the emf by a phase angle of π/2, we can write i c = e X t C 0 sin ffi fl fi ffffl / 2 ∴ The total current i in the circuit at this instant is i = i c + i L = e X t L 0 sin ffi fl fi ffffl / 2 + e XC 0 sin t ffi fl fi ffffl / 2 = e X t L 0 (ficosff )+ e XC 0 cos t ω = e0 cosωtfi( 1 1 X X c L ff–ff ff) i = e0 cosωtfi( ω ω C L fi–fi fi) 1 We find that, i = minimum when ω ω C L fi–fi 1 = 0 i.e. fi fi C L ff 1 i.e. ω2 = 1 LC ∴ ω = 1 LC or 2π fr = 1 LC ∴ fr = 1 2fflπ LC Where fr is called the resonant frequency. Therefore at parallel resonance frequency fr , i = minimum i.e. the circuit allows minimum current to flow through it. (as shown in the graph 13.20). Impedance is maximum at this frequency. The circuit is called parallel resonance circuit . A parallel resonant circuit is very useful in wireless transmission or radio communication and filter circuits. Characteristics of parallel resonance circuit 1. Resonance occurs when XL = XC. 2. Resonant frequency fr = 1 2ffl LC 3. Impedance is maximum 4. Current is minimum. 5. When alternating current of different frequencies are sent through parallel resonant circuit, it offers a very high impedance to the current of the resonant frequency ( fr ) and rejects it but allows the current of the other frequencies to pass through it, hence called a rejector circuit. Fig. 13.20: Parallel resonant curve. Resonance occurs in a series LCR circuit when XL = XC or fi ff 1 LC . For resonance to occur, the presence of both L and C elements in the circuit is essential. Only then the voltages L and C (being 180° out of phase) will cancel each other and current amplitude will be e0 /R i.e., the total source voltage will appear across. So we cannot have resonance LR and CR circuit. 13.9 Sharpness of Resonance: Q factor We have seen in section 13.4 (d) that the amplitude of current in the series LCR circuit is given by i 0 = e R L C 0 2 2 1 ffiffiffi ffi ffi fi ff ( ffi ffl ) ffl Also if ω is varied, then at a particular frequency ω =ω r , XL = XC i.e. ω r L= 1 ωrC .For a given resistance R, the amplitude of current is maximum when ω r L – 1 ωrC = 0 ∴ ω r = 1 LC For values of ω other than ωr , the amplitude of the current is less than the maximum value i0. Suppose we choose a value for ω for which the amplitude is 1 2 times its maximum value, the power dissipated by the circuit becomes half (called half power frequency).
303 Do you know? From the curve in the Fig. (13.21) we see that there are two such values of ω say ω 1 and ω 2 , one greater and other smaller than ω r and symmetrical about ω r such that ω 1 = ω r + Δω ω 2 = ω r – Δω The difference ω 1 – ω 2 = 2Δω is called the bandwidth of the circuit. The quantity ( fi fffi r 2 ) is regarded as the measure of the sharpness of resonance. The sharpness of resonance is measured by a coefficient called the quality or Q factor of the circuit The Q factor of a series resonant circuit is defined as the ratio of the resonant frequency to the difference in two frequencies taken on both sides of the resonant frequency such that at each frequency the current amplitude becomes 1 2 times the value at resonant frequency. ∴ Q= fi fi fi fi fffi r r Resonant frequency Bandwidth 2 1 ffl 2 ffi ffi Q-factor is a dimensionless quantity. The larger the value of Q-factor, the smaller the value of 2fiff or the bandwidth and sharper is the peak in the current or the series resonant circuit is more selective. Fig. (13.21) shows that the lower angular frequency side of the resonance curve is dominated by the capacitor’s reactance, the high angular frequency side is dominated by the inductor’s reactance and resonance occurs in the middle. Fig. 13.21: Sharpness resonance. The tuning circuit of a radio or TV is an example of LCR resonant circuit. Signals are transmitted by different stations at different frequencies which are picked up by the antenna. Corresponding to these frequencies a number of voltages appear across the series LCR circuit. But maximum current flows through the circuit for that AC voltage which has frequency equal to fr = 1 2ffl LC . If Q-value of the circuit is large, the signals of the other stations will be very weak. By changing the value of the adjustable capacitor C, the signal from the desired station can be tuned in. 13.10 Choke Coil: If we use a resistance to reduce the current passing through an AC circuit, there will be loss of electric energy in the form of heat (I 2 RT) due to Joule heating. A choke coil helps to minimise this effect. A choke coil is an inductor, used to reduce AC passing through a circuit without much loss of energy. It is made up of thick insulated copper wires wound closely in a large number of turns over a soft iron laminated core. Choke coil offers large resistance XL = ω L to the flow of AC and hence current is reduced. Laminated core reduces eddy current loss. Average power dissipated in the choke is P = I E rms rms cosφ , where the power factor cosφ = R R L 2 2 2 fiff . For a choke coil, L is very large. Hence R is very small so cosφ is nearly zero and power loss is very small. The only loss of energy is due hysteresis loss in the iron core, which can be reduced using a soft iron core. Internet my friend 1. https://en.m.wikipedia.org 2. hyperphysics.phy-astr.gsu.edu 3. https://www.britannica.com/science 4. www.khanacademy.org
304 1. Choose the correct option. i) If the rms current in a 50 Hz AC circuit is 5A, the value of the current 1/300 seconds after its value becomes zero is (A) 5 2 A (B) 5 3 2 A (C) 5 6 A D) 5 2 A ii) A resistor of 500 Ω and an inductance of 0.5 H are in series with an AC source which is given by V = 100 2 sin (1000 t). The power factor of the combination is (A) 1 2 (B) 1 3 (C) 0.5 (D) 0.6 iii) In a circuit L, C & R are connected in series with an alternating voltage of frequency f. the current leads the voltage by 450 . The value of C is (A) 1 fi fi ffl ffl f f ffl ffi 2 L Rff (B) 1 2 2 fi fi ffl ffl f f ffl ffi L Rff (C) 1 fi fi ffl ffl f f ffl ffi 2 L Rff (D) 1 2 2 fi fi ffl ffl f f ffl ffi L R ff iv) In an AC circuit, e and i are given by e = 150 sin (150t) V and i = 150 sin (150 t + π 3 ) A. the power dissipated in the circuit is (A) 106W (B) 150W (C) 5625W (D) Zero v) In a series LCR circuit the phase difference between the voltage and the current is 450 . Then the power factor will be (A) 0.607 (B) 0.707 (C) 0.808 (D) 1 2. Answer in brief. i) An electric lamp is connected in series with a capacitor and an AC source is glowing with a certain brightness. How does the brightness of the lamp change on increasing the capacitance ? ii) The total impedance of a circuit decreases when a capacitor is added in series with L and R. Explain why ? iii) For very high frequency AC supply, a capacitor behaves like a pure conductor. Why ? iv) What is wattles current ? v) What is the natural frequency of L C circuit ? What is the reactance of this circuit at this frequency 3. In a series LR circuit XL = R and power factor of the circuit is P1 . When capacitor with capacitance C such that XL = XC is put in series, the power factor becomes P2 . Calculate P1 / P2 . 4. When an AC source is connected to an ideal inductor show that the average power supplied by the source over a complete cycle is zero. 5. Prove that an ideal capacitor in an AC circuit does not dissipate power 6. (a) An emf e = e0 sin ωt applied to a series L - C – R circuit derives a current I = I 0 sinωt in the circuit. Deduce the expression for the average power dissipated in the circuit. (b) For circuits used for transporting electric power, a low power factor implies large power loss in transmission. Explain 7. A device Y is connected across an AC source of emf e = e0 sinωt. The current through Y is given as i = i 0 sin(ωt + π/2) a) Identify the device Y and write the expression for its reactance. b) Draw graphs showing variation of emf and current with time over one cycle of AC for Y. Exercises
305 c) How does the reactance of the device Y vary with the frequency of the AC ? Show graphically d) Draw the phasor diagram for the device Y. 8. Derive an expression for the impedance of an LCR circuit connected to an AC power supply. 9. Compare resistance and reactance. 10. Show that in an AC circuit containing a pure inductor, the voltage is ahead of current by π/2 in phase. 11. An AC source generating a voltage e = e0 sinωt is connected to a capacitor of capacitance C. Find the expression for the current i flowing through it. Plot a graph of e and i versus wt. 12. If the effective current in a 50 cycle AC circuit is 5 A, what is the peak value of current? What is the current 1/600 sec. after if was zero ? [Ans: 7.07A, 3.535 A] 13. A light bulb is rated 100W for 220 V AC supply of 50 Hz. Calculate (a) resistance of the bulb. (b) the rms current through the bulb. [Ans: 484Ω, 0.45A] 14. A 15.0 µF capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the current (rms and peak) in the circuit. If the frequency is doubled, what will happen to the capacitive reactance and the current. [Ans: 212Ω, 1.04 A, 1.47A, halved, doubled] 15. An AC circuit consists of only an inductor of inductance 2 H. If the current is represented by a sine wave of amplitude 0.25 A and frequency 60 Hz, calculate the effective potential difference across the inductor (π = 3.142) [Ans: 133.32V] 16. Alternating emf of e = 220 sin 100 πt is applied to a circuit containing an inductance of (1/π) henry. Write an equation for instantaneous current through the circuit. What will be the reading of the AC galvanometer connected in the circuit? [Ans: i = 2.2 sin (100π t-π/2), 1.555A] 17. A 25 µF capacitor, a 0.10 H inductor and a 25Ω resistor are connected in series with an AC source whose emf is given by e = 310 sin 314 t (volt). What is the frequency, reactance, impedance, current and phase angle of the circuit? [Ans: 50Hz, 95.9Ω, 99.1Ω, 2.21A, 1.31 rad] 18. A capacitor of 100 µF, a coil of resistance 50Ω and an inductance 0.5 H are connected in series with a 110 V-50Hz source. Calculate the rms value of current in the circuit. [Ans: 0.816A] 19. Find the capacity of a capacitor which when put in series with a 10Ω resistor makes the power factor equal to 0.5. Assume an 80V-100Hz AC supply. [Ans: 9.2 × 10-5 F] 20. Find the time required for a 50 Hz alternating current to change its value from zero to the rms value. [Ans: 2.5 × 10-3 s] 21. Calculate the value of capacity in picofarad, which will make 101.4 micro henry inductance to oscillate with frequency of one megahertz. [Ans: 249.7 picofarad] 22. A 10 µF capacitor is charged to a 25 volt of potential. The battery is disconnected and a pure 100 m H coil is connected across the capacitor so that LC oscillations are set up. Calculate the maximum current in the coil. [Ans: 0.25 A] 23. A 100 µF capacitor is charged with a 50 V source supply. Then source supply is removed and the capacitor is connected across an inductance, as a result of which 5A current flows through the inductance. Calculate the value of the inductance. [Ans: 0.01 H]
306 Can you recall? 14.1 Introduction: In earlier chapters you have studied various optical phenomena like reflection, refraction, interference, diffraction and polarization of light. Light is electromagnetic radiation and most of the phenomena mentioned have been explained considering light as a wave. We are also familiar with the wave nature of electromagnetic radiation in other regions like X-rays, γ-rays, infrared and ultraviolet radiation and microwaves apart from the visible light. Electromagnetic radiation consists of mutually perpendicular oscillating electric and magnetic fields, both being perpendicular to the direction in which the wave and energy are travelling. In Chapter 3 on Kinetic Theory of Gases and Radiation, you have come across spectrum of black body radiation which cannot be explained using the wave nature of radiation. Such phenomena appear during the interaction of radiation with matter and need quantum physics to explain them. The idea of 'quantization of energy' was first proposed by Planck to explain the black body spectrum. Planck proposed a model that says (i) energy is emitted in packets and (ii) at higher frequencies, the energy of a packet is large. Planck assumed that atoms behave like tiny oscillators that emit electromagnetic radiation only in discrete packets (E = nhν), where ν is the frequency of oscillator. The emissions occur only when the oscillator makes a jump from one quantized level of energy to another of lower energy. This model of Planck turned out to be the basis for Einstein’s theory to explain the observations of experiments on photoelectric effect which we will study in the following section. 14.2 The Photoelectric Effect: Heinrich Hertz discovered photoelectric emission in 1887 while he was working on the production of electromagnetic waves by spark discharge. He noticed that when ultraviolet light is incident on a metal electrode, a high voltage spark passes across the electrodes. Actually electrons were emitted from the metal surface. The surface which emits electrons, when illuminated with appropriate radiation, is known as a photosensitive surface. 1. What is electromagnetic radiation? 2. What are the characteristics of a wave? 3. What do you mean by frequency and wave number associated with a wave? 4. What are the characteristic properties of particles of matter? 5. How do we define momentum of a particle? 6. What are the different types of energies that a particle of matter can possess? 14. Dual Nature of Radiation and Matter Fig. 14.1: Process of photoelectric effect. The phenomenon of emission of electrons from a metal surface, when radiation of appropriate frequency is incident on it, as shown in Fig. 14.1, is known as photoelectric effect. For metals like zinc, cadmium, magnesium etc., ultraviolet radiation is necessary while for alkali metals, even visible radiation is sufficient. Electrical energy can be obtained from light (electromagnetic radiation) in two ways (i) photo-emissive effect as described above and (ii) photo-voltaic effect, used in a solar cell. In the latter case, an electrical potential difference is generated in a semiconductor using solar energy.
307 14.2.1 Experimental Set-up of Photoelectric Effect: A typical laboratory experimental set-up for the photoelectric effect (Fig. 14.2) consists of an evacuated glass tube with a quartz window containing a photosensitive metal plate - the emitter E and another metal plate - the collector C. The emitter and collector are connected to a voltage source whose voltage can be changed and to an ammeter to measure the current in the circuit. A potential difference of V, as measured by the voltmeter, is maintained between the emitter E (the cathode) and collector C (the anode), normally C being at a positive potential with respect to the emitter. This potential difference can be varied and C can even be at negative potential with respect to E. When the anode potential V is positive, it accelerates the electrons (hence called accelerating potential) while when the anode potential V is negative, it retards the flow of electrons (therefore known as retarding potential). A source S of monochromatic light (light corresponding to only one specific frequency) of sufficiently high frequency (short wavelength ≤ 10-7 m) is used. from the metal through its surface. These electrons, called photoelectrons, are collected at the collector C (photoelectron are ordinary electrons, they are given this name to indicate that they are emitted due to incident light). We now know that free electrons are available in a metal plate. They are emitted if sufficient energy (we will know more about this energy later in the Chapter) is supplied to them to overcome the barrier that keeps them inside the metal. In the late nineteenth century, these facts were not known and scientists working on photoelectric effect performed various experiments and noted down their observations. These observations are summarized below. We will try to analyze these observations and their explanation. 14.2.2 Observations from Experiments on Photoelectric Effect: 1. When ultraviolet radiation was incident on the emitter plate, current I was recorded even if the intensity of radiation was very low. Photocurrent I was observed only if the frequency of the incident radiation was more than some threshold frequency ν0 . ν0 was same for a given metal and was different for different metals used as the emitter. For a given frequency ν ( > ν0 ) of the incident radiation, no matter how feeble was the light meaning however small the intensity of radiation be, electrons were always emitted. 2. There was no time lag between the incidence of light and emission of electrons. The photocurrent started instantaneously (within 10-9 s) on shining the radiation even if the intensity of radiation was low. As soon as the incident radiation was stopped, the flow of current stopped. 3. Keeping the frequency ν of the incident radiation and accelerating potential V fixed, if the intensity was increased, the photo current increased linearly with intensity as shown in Fig. 14.3. Fig. 14.2: Schematic of experimental set-up for photoelectric effect. Light is made to fall on the surface of the metal plate E and electrons are ejected
308 4. The photocurrent I could also be varied by changing the potential of the collector plate. I was dependent on the accelerating potential V (potential difference between the emitter and collector) for given incident radiation (intensity and frequency were fixed). Initially the current increased with voltage but then it remained constant. This was termed as the saturation current I 0 (Fig. 14.4). Fig. 14.3: Photocurrent as a function of incident intensity for fixed incident frequency and accelerating potential . 5. Keeping the accelerating voltage and incident frequency fixed, if the intensity of incident radiation was increased, the value of saturation current also increased proportionately, e.g., if the intensity was doubled, the saturation current was also doubled. 6. The maximum kinetic energy KEmax (and hence the maximum velocity) of the electrons depended on the potential V for a given metal used for the emitter plate and for a given frequency of the incident radiation. If the material is changed or Fig. 14.4: Photocurrent as a function of accelerating potential for fixed incident frequency and different incident intensities. the frequency of the incident radiation is changed, KEmax changed. It did not depend on the intensity of the incident radiation. Thus, even for very small incident intensity, if the frequency of incident radiation was larger than the threshold frequency v0 , KEmax from a given surface was always the same for a given incident frequency. 7. If increasingly negative potentials were applied to the collector, the photocurrent decreased and for some typical value -V0 , photocurrent became zero. V0 was termed as cut-off or stopping potential. It indicated that when the potential was retarding, the photoelectrons still had enough energy to overcome the retarding (opposing) electric field and reach the collector. Value of V0 was same for any incident intensity as long as the incident frequency was same (Fig. 14.4) but was different for different emitter materials. 8. If the frequency of incident radiation was changed keeping the intensity and accelerating potential V constant, then the saturation current remained the same but the stopping potential V0 changed. This observation is depicted in Fig. 14.5. The stopping potential V0 varied linearly with ν as shown in Fig. 14.6. For different metals, the slopes of such straight lines were the same but the intercepts on the frequency and stopping potential axes were different. Fig. 14.5: Photocurrent as a function of accelerating potential for fixed incident intensity but different incident frequencies for the same emitter material . v1 v2 v3 v3 > v2 > v1
309 You have studied ionization energy of an atom. What is ionization energy to an atom is the work function to a solid which is a large collection of atoms. We know that metals have free electrons. This fact makes metals good conductors of heat and electricity. These electrons are free to move inside the metal but are otherwise confined inside the metal. They cannot escape from the surface unless sufficient energy is supplied to them. The minimum amount of energy required to be provided to an electron to pull it out of the metal from the surface is called the work function of the metal and is denoted by φ0 . Work function depends on the properties of the metal and the nature of its surface. Values of work function of metals are generally expressed in a unit of energy called the electron volt (eV). 9. The photocurrent and hence the number of electrons depended on the intensity but not on the frequency of incident radiation, as long as the incident frequency was larger than the threshold frequency ν0 and the potential of anode was higher than that of cathode. 14.2.3 Failure of Wave Theory to Explain the Observations from Experiments on Photoelectric Effect: Most of these observations could not be explained by the wave theory of electromagnetic radiation. First and foremost was the instantaneous emission of electrons on incidence of light. Wave picture would expect that the metal surface will absorb the incident energy continuously. All the electrons near the surface will absorb energy. The metal surface will require reasonable time (~ few minutes to hours) to accumulate sufficient energy to knock off electrons. Greater the intensity of incident radiation, more will be the incident energy, hence expected time required to knock off the electrons will be less. For small incident intensity, the energy incident on unit area in unit time will be small, and will take longer to knock off the electrons. These arguments were contradictory to observations. Let us try to estimate the time that will be required for the photocurrent to start. We need to define the term ‘work function’ of a metal for this exercise. Fig. 14.6: Stopping potential as a function of frequency of incident radiation for emitters made of different metals. Example 14.1: Radiation of intensity 0.5 × 10-4 W/m2 falls on the emitter in a photoelectric set-up. The emitter (cathode) is made up of potassium and has an area of 5 cm2 . Let us assume that the electrons from only the surface are knocked off by the radiation. According to the wave theory, what will be the time required to notice some deflection in the microammeter Table 14.1 : Typical values of work function for some common metals. Metal Work function (in eV) Potassium 2.3 Sodium 2.4 Calcium 2.9 Zinc 3.6 Silver 4.3 Aluminum 4.3 Tungsten 4.5 Copper 4.7 Nickel 5.0 Gold 5.1
310 maximum kinetic energy did not depend on the incident intensity but depended on the incident frequency. According to wave theory, frequency of incident radiation has no role in determining the kinetic energy of photoelectrons. Moreover, wave theory expected photoelectrons to be emitted for any frequency if the intensity of radiation was large enough. But observations indicated that for a given metal surface, some characteristic cut-off frequency ν0 existed below which no photoelectrons were emitted however intense the incident radiation was and photoelectrons were always emitted if incident frequency ν was greater than ν0 even if the intensity was low. 14.2.4 Einstein’s Postulate of Quantization of Energy and the Photoelectric Equation: Planck’s hypothesis of energy quantization to explain the black body radiation was extended by Einstein in 1905 to all types of electromagnetic radiations. Einstein proposed that under certain conditions, light behaves as if it was a particle and its energy is released or absorbed in bundles or quanta. He named the quantum of energy of light as photon with energy E = hν, where ν is the frequency of light and h is a constant defined by Planck in his model to explain black body radiation. It is now known as the Planck’s constant and has a value 6.626 × 10-34 J s. It may be noted that the equation E = hν --- (14.1) is a relation between a particle like property, the energy E and a wave like property, the frequency ν. Equation (14.1) is known as the Einstein’s relation. Einstein’s relation (14.1) holds good for the entire electromagnetic spectrum. It says that energy of electromagnetic radiation is directly proportional to the frequency (and is inversely proportional to the wavelength since ν = c/λ). Hence high frequency radiation connected in the circuit? (Given the metallic radius of potassium atom is 230 pm and work function of potassium is 2.3 eV.) Solution : Given Intensity of radiation = 0.5 × 10-4 W/m2 , Area of cathode = 5 cm2 = 5 × 10-4 m2 . Radius of potassium atom = 230 pm = 230 × 10-12 m Work function of potassium = 2.3 eV = 2.3 × 1.6 ×10-19 J The number N of electrons present on the surface of cathode can be approximately calculated assuming that each potassium atom contributes one electron and the radius of potassium atom is 230×10-12 m. N = Area of cathode/ area covered by one atom = 5×10-4/(3.1415×230×10-12×230×10-12) = 3009×1012 Incident power on the cathode is = 0.5 × 10-4 W/m2 × 5×10-4 m2 = 2.5×10-8 W Wave theory assumes that this power distributed over the whole area of the cathode is uniformly absorbed by all the electrons. Therefore the energy absorbed by each electron in one second is = 2.5×10-8 W /3009×1012 ≈ 8.31×10-24 W. Work function of potassium is 2.30 eV = 2.30 × 1.6 ×10-19 J = 3.68 ×10-19 J. Hence each electron will require minimum 3.68×10-19 J of energy to be knocked off from the surface of the cathode. The time required to accumulate this energy will be 3.68 × 10-19 J / 8.31 × 10-24 W = 0.443 × 105 s, which is about half a day. Secondly, since larger incident intensity implies larger energy, the electrons are expected to be emitted with larger kinetic energy. But the observation showed that the
311 Try this means high energy radiation. Alternatively, short wavelength radiation means high energy radiation. Example 14.2: (a) Calculate the energies of photons corresponding to ultraviolet light and red light, given that their wavelengths are 3000 Å and 7000 Å respectively. (Remember that the photon are not coloured. Colour is human perception for that frequency range.) (b) A typical FM radio station has its broadcast frequency 98.3 MHz. What is the energy of an FM photon of this frequency? Solution: Given λuv = 3000 Å = 3000 × 10-10 m, λred = 7000 Å = 7000 × 10-10 m and νFM = 98.3 MHz = 98.3 × 106 s-1 We know that energy E of electromagnetic radiation of frequency ν is hν and if λ is the corresponding wavelength, then λν = c, c being the speed of electromagnetic radiation in vacuum. Hence, E h hc fi fi ff ffl (a) E fi ff ff ff ff fi ffl ffl ffl 6 63 10 3 10 3000 10 6 6 34 8 1 10 . . Js ms m 3 10 19 ff ffl J = 4.147 eV for a photon corresponding to ultraviolet light and E fi ff ff ff ff fi ffl ffl ffl 6 63 10 3 10 7000 10 2 84 34 8 1 10 . . Js ms m ff ffl 10 19 J = 1.77 eV for a photon corresponding to red light. (b) The energy of photon of FM frequency 98.3 MHz is 6.63 ×10-34 J s × 98.3 × 106 s-1 = 651.73 ×10-28 J = 40.73 ×10-8 eV. This is very small energy as compared to the photon energy in the visible range. The explanation using Einstein's postulate of quantization of energy for the observations mentioned in section 14.2.2 is given below. 1. Einstein argued that when a photon of ultraviolet radiation arrives at the metal surface and collides with an electron, it gives all of its energy hν to the electron. The energy is gained by the electron and the photon no longer exists. If φ0 is the work function of the material of the emitter plate, then electrons will be emitted if and only if the energy gained by the electrons is more than or equal to the work function i.e.,hν ≥ φ0 . Thus, a minimum or threshold frequency ν0 (= φ0 /h) is required to eject electrons from the metal surface. If ν < ν0 , the photon will not have enough energy to liberate an electron. As a result, no electron will be ejected however intense the incident radiation is. Similarly if ν > ν0 ,the energy will always be sufficient to eject an electron, however small the incident intensity is. 2. Energy is given by the photon to the electron as soon as the radiation is incident on the surface. The exchange of energy between the photon and electron • Wavelength (in Å) × energy (in eV) ≈ 12500 (numerically) • Wavelength (in nm) × energy (in eV) ≈ 1250 (numerically) Determine the wavelengths and frequencies for photons of energies (i) 10-12 J, (ii) 10-15 J, (iii) 10-18 J, (iv) 10-21 J and (v) 10-24 J. Accordingly prepare a chart (along a horizontal line) of various regions of electromagnetic spectrum and identify these regions in categories that you know. Compare your results with a standard chart from any reference book or from Internet. You would notice that γ photons are the most energetic photons and their energies are ~ 10-13 - 10-12 J. This is a very small amount of energy on the human scale and therefore we do not notice individual photons along their passage.
312 6. If the frequency of incident radiation is more than the threshold frequency, then the energy φ0 is used by the electron to escape from the metal surface and remaining energy of the photon becomes the kinetic energy of the electron. Depending on the energy of the electron inside the metal and other processes like collisions after emission from the surface, the maximum kinetic energy is equal to (hν - φ0 ). Hence, KEmax = hν - φ0 --- (14.2) Equation (14.2) is known as Einstein’s photoelectric equation. KEmax depends on the material of the emitter plate and varies linearly with the incident frequency ν; it is independent of the intensity of the incident radiation. 7. The electrons that are emitted from the metal surface have different kinetic energies. The reasons for this are manyfold: all the electrons in a solid do not possess the same energy, the electrons may be ejected from varying depths inside the metal surface, electrons may suffer collisions before they come out of the metal surface and may lose their energy etc. If V is the potential difference between the emitter and collector and the collector is at a lower potential, an electron will lose its kinetic energy in overcoming the retarding force. If the kinetic energy is not sufficient, the emitted electrons may not reach the collector and the photocurrent will be zero. If KEmax is the energy of the most energetic electron at the emitter surface (where its potential energy is zero) and -V0 is the stopping potential, then this electron will fail to reach the collector if KEmax< eV0, where e is the electron charge and eV0 is the energy needed for the electron to overcome the retarding potential V0 . If the electron just fails to is instantaneous. Hence there is no time lag between the incidence of light and emission of electrons. Also when the incident radiation is stopped, there are no photons to transfer the energy to electrons, hence the photoemission stops immediately. 3. According to Einstein’s proposition, if the intensity of incident radiation for a given wavelength is increased, there will be an increase in the number of energy quanta (photons) incident on unit area in unit time; the energy of each quantum being the same (= hν = hc/λ). Therefore larger intensity radiation will knock off more number of electrons from the surface and hence the current will be larger (if ν > ν0 ). Conversely lower intensity implies less number of incident photons, hence, less number of ejected electrons and therefore lower current. 4. Once the electron is emitted from the surface, if the collector is at a higher potential than the emitter, the electric field will accelerate the electrons towards the collector. Higher is the accelerating potential, more will be number of electrons reaching the collector. Hence the photocurrent I increases with the accelerating potential initially. Moreover, since the intensity of incident radiation determines the number of photons incident on the metal surface on unit area in unit time, it determines the maximum number of electrons that can be knocked off by the incident radiation. Hence for a given intensity, increasing the accelerating potential can increase the current only till all the knocked off electrons have reached the collector. No increase can be seen in the current beyond this limit. This explains the saturation current I0 . 5. Increasing the incident intensity will increase the number of incident photons and eventually the saturation current.
313 Use your brain power was accepted. The work function values φ0 for some metals were also confirmed from Eq. (14.3). Einstein and Millikan received Nobel prizes for their respective discoveries in 1921 and 1923 respectively. reach the collector, i.e., it has lost all its kinetic energy just at the collector, KEmax= eV0 and the photocurrent becomes zero. Equation (14.2) then explains that stopping potential V0 depends on the incident frequency and the material of the emitter and does not depend on the incident intensity. 8. If the ejected electrons have kinetic energy more than eV0 , electrons can reach the collector, hence current flows. When the kinetic energy of the electron is less than or equal to eV0 , no current will flow. Photocurrent will become zero when KEmax = eV0 . Using KEmax = eV0 , we can write Eq. (14.2) as eV h 0 0 ffff ffff fi ff ffl ffi or, V h e e 0 0 fi ffff ff ffl ffi fl --- (14.3) Above equation tells us that V0 varies linearly with incident frequency ν, and the slope of the straight line depends on constants h and e while the intercept of the line depends on the material through φ0 . Thus the slope of lines in Fig. 14.6 is same and is independent of the material of the emitter but intercepts are different for different materials. 9. All the above arguments thus bring out the fact that the magnitude of photocurrent depends on the incident intensity through the number of emitted photoelectrons and the potential V of the collector but not on the incident frequency ν as long as ν > ν0 . Thus all the observations related to the experiments on photoelectric effect were explained by Einstein’s hypothesis of existence of a photon or treating light as bundles of energy. Although Einstein gave his hypothesis in 1905, it was not widely accepted by the scientific community. In 1909, when Millikan measured the charge of an electron and the value of h, calculated from Eq. (14.3), matched with the value given by Planck, the hypothesis You must have seen light emitting diodes (LEDs) of different colours. In LED, electrical energy is converted into light energy corresponding to different colours. Can you tell what must be the difference in the working of LEDs of different colours. Design an experiment using LEDs to determine the value of Planck’s constant. You might know that Nobel prize in physics for the year 2014 was awarded to Professors Isamu Akasaki, Hiroshi Amano and Shuji Nakamura for the invention of blue LEDs. They made the first blue LED in the early 1990s. Try to search on the Internet why it was difficult to make a blue LED. Example 14.3: The wavelength and intensity of the incident light is 4000 Å and 0.1 W respectively. What is the minimum change in the light energy? What is the number of incident photons? Solution : Given incident intensity = 0.1 W and λ = 4000 Å = 4000 × 10-10 m. The energy E of a photon of given wavelength is E h hc fi fi fi ff ff ff ff ffl ffl ffi fl fi fi . fi / fifi 6 63 10 3 10 4000 10 34 8 10 Jfisfi fi fim s m fi ff ffl 4 97 10 19 . fi fiJ This is the minimum change in energy and is very small. The change in energy can therefore be considered as continuous. According to Einstein, energy of radiation of frequency ν comes in bundles with magnitude hν. Thus energy of a light beam having n photons will be nhν, where n can take only integral values. Is it then possible to vary the incident energy continuously? Why we do not see individual photons? To understand this issue, let us consider the following example.
314 Can you tell? Table 14.2 : Summary of analysis of observations from experiments on photoelectric effect. Observation Wave theory Photon picture Electrons are emitted as soon as the light is incident on the metal surface. Very intense light is needed for instantaneous emission of electrons. Only one photon is needed to eject one electron from the metal surface and energy exchange between electron and photon is instantaneous on collision. Very low intensity of incident light is also sufficient to generate photocurrent. Low intensity should not give photocurrent. Low intensity of incident light means less number of photons and not low energy photons. Hence low current will be produced. High intensity gives larger photocurrent means higher rate of release of electrons. High intensity means higher energy radiation and therefore more electrons are emitted. Higher intensity means more number of photons incident in unit time, therefore more number of electrons are emitted in unit time and hence photocurrent is larger. Increasing the intensity has no effect on the electron energy. Higher intensity should mean electrons emitted with higher energies. Higher intensity means higher number of incident photons per unit time. Energy of photon is same as it does not depend on the intensity. A minimum threshold frequency is needed for photocurrent to start. Low frequency light should release electrons but would take more time. A photon of low frequency light will not have sufficient energy to release an electron from the surface. Increasing the frequency of incident light increases the maximum kinetic energy of electrons. Increasing intensity should increase the maximum kinetic energy. Maximum kinetic energy should not depend on the incident frequency. Increasing the frequency increases the energy of the photon. Therefore electrons receive more energy which results in increasing the maximum kinetic energy. Number of photons N incident per second is N fi ff ffl ff ffi . fi . fi 0 1 4 97 10 2 10 19 W 17 J The number of photons coming out is so large that human eye cannot comprehend or count it. Even if one wishes to count, say 10 photons per second, ∼109 years will be required. A particular metal used as a cathode in an experiment on photoelectric effect does not show photoelectric effect when it is illuminated with green light. Which of the colours in the visible spectrum are likely to generate photocurrent? 14.3 Wave-Particle Duality of Electromagnetic Radiation: In its interaction with matter, light behaves as if it is made up of packets of energy called quanta. Later it was confirmed from other theoretical and experimental investigations that these light quanta can have associated momentum. Hence the question came up whether a particle can be associated with light or electromagnetic radiation in general. Particle nature was confirmed by Compton in 1924 in experiments on scattering of X-rays due to electrons of matter. Summary of these results is given in the box below and you can
315 Do you know? Compton showed that photon has an associated momentum along with the energy it carries. All photons of electromagnetic radiation of a particular frequency have the same energy and momentum. Photons are electrically neutral and are not deflected by electric or magnetic fields. Photons can have particle-like collisions with other particles such as electrons. In photon – particle collision, energy and momentum of the system are conserved but the number of photons is not conserved. Photons can be absorbed or new photons can be created. Photons can transfer their energy and momentum during collisions with particles and disappear. When we turn on light, they are created. Photon always moves with the speed of light, it is never at rest. Mass of a photon is not defined as we do for a particle in Newtonian mechanics. Its rest mass is zero (in all frames of reference). Effects of wave nature of light were seen in experiments on interference or diffraction when the slit widths or the separation between two slits are smaller than or comparable to the wavelength of light. If the slit width is large or the spacing between slits is more, the interference or diffraction patterns will not be same and the wave nature will not be so obvious. It was realized by scientists that some phenomena observed in experiments in the laboratory or in nature (like interference and know more about these experiments from the reference books given at the end of this book or from the links given below • http://physics.usask.ca/~bzulkosk/ modphyslab/phys251manual/ compton_2009.pdf • http://www.phys.utk.edu/labs/modphys/ Compton Scattering Experiment.pdf • http://hyperphysics.phy-astr.gsu.edu/ hbase/quantum/comptint.html The particle nature of radiation is seen in black body radiation and photoelectric effect. In the former, near room temperature, the radiation is mostly in the infrared region while in the latter it is in the visible and ultraviolet region of the spectrum. The third experiment, which established that a photon possesses momentum like a particle, was Compton scattering where X-rays and γ-rays interact with matter. In 1923, A. H. Compton made a monochromatic beam of X-rays, of wavelength λ, incident on a graphite sheet and measured the intensity of the scattered rays in different directions as a function of wavelength. He found that although the incident beam consisted of a single wavelength λ, the scattered intensity was maximum at two wavelengths. One of these was same as the incident wavelength but the other λ′ was larger by an amount ∆λ. ∆λ is known as the Compton shift that depends on the scattering angle. Compton explained his observations by considering incidence of X-ray beam on graphite as collision of X-ray photons with the electrons of graphite, like collision of billiard balls. Energy and momentum is transferred during the collision and scattered photons have lower energy than the incident photons. Therefore they have lower frequency or higher wavelength. The Compton shift is given by the relation fiff ff fi fi ffiflffi ' ff ffl fi fi ( c fl os ) h m ce 1 where θ is the scattering angle. The shift depends only on the scattering angle and not on the incident wavelength. This shift cannot be explained using wave theory. If we let the Planck’s constant go to zero, we get the result expected from wave theory. This is the test to check whether the new picture is correct or not.
316 Use your brain power diffraction) can be explained by considering light in particular, and electromagnetic radiation in general, as a wave. On the other hand, some other observations (like photoelectric effect and black body radiation) can be explained only if we consider electromagnetic radiation as consisting of photons with definite quantum of energy (and momentum as evident from Compton scattering experiments). Also there are some phenomena which can be explained by both the theories. It is therefore essential to consider that both the characters or behaviours hold good; one dominates in some situations and the other works in rest of the situations. It is necessary to keep both the physical models to explain the careful experimental observations. There is thus a need to hypothesize the dual character of light. Later it turned out that such a picture is required not only for light but for the whole electromagnetic spectrum. This phenomenon is termed as wave-particle duality of electromagnetic radiation. 14.4 Photo Cell: Photo cell is a device that makes use of the photoelectric effect and converts light energy into electrical energy. Schematic of a photocell is shown in Fig. 14.7. It consists of a semi-cylindrical photosensitive metal plate E (acting as a cathode) and a wire loop collector C (acting as an anode) supported in an evacuated glass or quartz bulb. The electrodes are connected to an external circuit having a high tension battery B and a microammeter µA. Instead of a photosensitive metal plate, the photosensitive material can be pasted in the form of a thin film on the inner walls of the glass bulb. When light of suitable wavelength falls on the cathode, photoelectrons are emitted. These electrons are attracted towards the anode due to the applied electric field. The generated photocurrent is noted from the microammeter. Photocell is used to operate control systems and in light measuring devices. Light meters in photographic cameras make use of photocell to measure the intensity of light. Photocell can also be used to switch on or off the street lights. Fig. 14.7 : Schematic of a photocell. Suppose source of ultraviolet radiation is kept near the passage or entrance of a mall or house and the light is made incident on the cathode of a photocell, photocurrent is generated. When a person passes through the passage or comes near the entrance, incident light beam is interrupted and photocurrent stops. This event can be used to operate a counter in counting devices, or to set a burglar alarm. Such an arrangement can be used to identify traffic law defaulters by setting an alarm using the photocell. Is solar cell a photocell? 14.5 De Broglie Hypothesis: In 1924, Prince Louis de Broglie (pronounced as ‘de broy’) proposed, on the basis of the symmetry existing in nature, that if radiation has dual nature - sometimes wave nature dominates and sometimes particle nature, matter may also possess dual nature. Normally we talk about matter as composed of particles, but are there situations where matter seems to show wave-like properties? This will become evident from the experiments on diffraction of electrons from nickel crystals described later in this chapter. De Broglie used the properties, frequency ν and wavelength λ, of a wave and proposed a relation to connect these with the particle
317 Can you tell? Use your brain power properties, energy E and momentum p. The momentum p carried by a photon of energy E is given by the relation p E c = --- (14.4) which is valid for a massless particle travelling with the speed of light c according to Einstein's special theory of relativity. Using the Einstein’s relation for E, p E c h c h fi fi fi ff ffl --- (14.5) where λ, the wavelength, is given by λν = c. De Broglie proposed that a moving material particle of total energy E and momentum p has associated with it a wave analogous to a photon. He then suggested that the wave and particle properties of matter can also be described by a relation similar to Eq. (14.5) for a photon. Thus frequency and wavelength of a wave associated with a material particle, of mass m moving with a velocity v, are given as ν = E/h and λ = h/p = h/mv --- (14.6) He referred to these waves associated with material particles as matter waves. The wavelength of the matter waves, given by Eq. (14.6), is now known as de Broglie wavelength. Greater is the momentum, shorter is the wavelength. Equation (14.6) for the wavelength of matter waves is known as de Broglie relation. For a particle of mass m moving with a velocity v, the kinetic energy EK = 1 2 mv2 or v = 2E m K . Thus, fi ff ff ff h m h m m E h K mEK v 2 2 For a charged particle of charge q, accelerated from rest, through a potential difference V, the work done is qV. This provides the kinetic energy. Thus EK = qV. fi ff ffl ff h mE h 2 2 K mqV . This relation holds for any charged particle like electron, proton or for even charged ions where m corresponds to the mass of the charged particle. Of course, when V is very large (say in kV), so that the speed of the particle becomes close to the speed of light, such an equation will not be applicable. You will learn about other effects in such situations in higher classes. For an electron moving through a potential difference of V (given in volts) fi ff h 2m ee V fi ff ff ff ff ff ff fi ffl ffl ffl ffi fl 6 63 10 2 9 11 10 1 6 10 1 34 31 19 . . . Js kg C i V nvolts .228 10 9 ff ffl V ffi fl involts m or, fifi fi . infinm infivolts ff ffl ffi ff ffl 1 228 V --- (14.7) Example 14.4: An electron is accelerated through a potential of 120 V. Find its de Broglie wavelength. Solution: Given V = 120 V. We know that λ = 1 228. V using Eq. (14.7). ∴ λ = 1 228 120 . = 0.112 nm. The expression p = E/c defines the momentum of a photon. Can this expression be used for momentum of an electron or proton? Can you estimate the de Broglie wavelength of the Earth? Shortly after the existence of photons (particles associated with electromagnetic waves) was postulated, it was also experimentally found that electrons sub-atomic and atomic particles like protons and neutrons
318 also exhibit wave properties. The wavelength associated with an electron of energy few eV is of the order of few Å. Therefore to observe the wave nature of electron, slit width or diffracting objects should be of same order of magnitude (few Å). The wave property of electron was confirmed experimentally in 1927 by Davisson and Germer in America and in 1928 by George P. Thomson in England by diffraction of electrons by atoms in metals. Knowing that the size of the atoms and their spacing in crystals is of the order of few Å, they anticipated that if electrons are scattered by atoms in a crystal, the associated matter waves will interfere and will show diffraction effects. It turned out to be true in their experiments. Electrons showed constructive and destructive interference. No electrons were found in certain directions due to destructive interference while in other directions, maximum numbers of electrons were seen due to constructive interference. Louis de Broglie received the Nobel prize in Physics in 1929 and Davisson, Germer and Thomson shared the Nobel prize in Physics in 1937. It was amazing that Sir J. J. Thomson discovered the existence of electron as a subatomic particle while his son G. P. Thomson showed that electron behaves like a wave. 14.6 Davisson and Germer Experiment: A schematic of the experimental arrangement of the Davisson and Germer experiment is shown in Fig. 14.8. The whole set-up is enclosed in an evacuated chamber. It uses an electron gun - a device to produce electrons by heating a tungsten filament F using a battery B. Electrons from the gun are accelerated through vacuum to a desired velocity by applying suitable accelerating potential across a cylindrical anode and are collimated into a focused beam. This beam of electrons falls on a nickel crystal and is scattered in different directions by the atoms of the crystal. Thus, in the Davisson and Germer experiment, electrons were used in place of light waves. Scattered electrons were detected by an electron detector and the current was measured with the help of a galvanometer. By moving the detector on a circular scale that is by changing the scattering angle θ (angle between the incident and the scattered electron beams), the intensity of the scattered electron beam was measured for different values of scattering angle. Scattered intensity was not found to be uniform in all directions (as predicted by classical theory). The intensity pattern resembled a diffraction pattern with peaks corresponding to constructive interference and troughs to regions of destructive interference. Diffraction is a property of waves. Hence, above observations implied that the electrons formed a diffraction pattern on scattering and that particles could show wave-like properties. Fig. 14.8: Schematic of Davisson and Germer experiment. Davisson and Germer varied the accelerating potential from 44 V to 68 V and observed a peak in the intensity of the scattered electrons at scattering angle of 50º for a potential of 54 V. This peak was the result of constructive interference of the electrons scattered from different layers of the regularly spaced atoms of the nickel crystal. From Eq. (14.7), for V = 54 V, we get λ = 1.228/√54 = 0.167 nm --- (14.8) From the electron diffraction measurements, the wavelength of matter waves associated with the electrons was found to be 0.165 nm. The two values of λ,
319 Use your brain power obtained from the experimental results and from the theoretical de Broglie relation, were in close agreement. The Davisson and Germer experiment thus substantiated de Broglie’s hypothesis of wave-particle duality and verified his relation. particles. Wave-particle duality implies that all moving particles have an associated frequency and an associated wave number and all waves have an associated energy and an associated momentum. We come across the wave-particle duality of matter due to quantum behaviour when we are dealing with microscopic objects (sizes ≤ 10-6 m). Small order of magnitude of h sets the scale at which quantum phenomena manifest themselves. If all the material objects in motion have an associated wavelength (and therefore an associated wave), why then we do not talk about wavelength of a child running with speed v on a pathway 2 m wide or a car moving with speed v on a road 20 m wide? To understand this, let us try to calculate these quantities. Diffraction results described above can be produced in the laboratory using an electron diffraction tube as shown in figure. It has a filament which on heating produces electrons. This filament acts as a cathode. Electrons are accelerated to quite high speeds by creating large potential difference between the cathode and a positive electrode. On its way, the beam of electrons comes across a thin sheet of graphite. The electrons are diffracted by the atomic layers in the graphite and form diffraction rings on the phosphor screen. By changing the voltage between the cathode and anode, the energy, and therefore the speed, of the electrons can be changed. This will change the wavelength of the electrons and a change will be seen in the diffraction pattern. By increasing the voltage, the radius of the diffraction rings will decrease. Try to explain why? Example 14.5: A student, weighing 45 kg, is running with a speed of 8 km per hr on a foot path 2 m wide. A small car, weighing 1200 kg, is moving with a speed of 60 km per hr on a 20 m wide road. Calculate their de Broglie wavelengths. Solution : Given v1 = 8 km / hr = 8 × 103 /3600 m / s and m1 = 45 kg for the student, v2 = 60 km / hr = 60 × 103 /3600 m / s and m2 = 1200 kg for the car, momentum p1 = 45 × 8 × 103 /3600 = 100 kg m /s for the student and momentum p2 = 1200 × 60 × 103 /3600 = 20000 kg m /s for the car. The de Broglie wavelength λ1 = h/p1 = 6.63 × 10-34 J s / 100 kg m /s = 6.63 × 10-36 m. for the student, and de Broglie wavelength λ2 = h/p2 = 6.63 × 10-34 J s/ 20000 kg m /s) = 3.32 × 10-38 m for the car. 14.7 Wave-Particle Duality of Matter: Material particles show wave-like nature under certain circumstances. This phenomenon is known as wave-particle duality of matter. Frequency ω and wave number k are used to describe waves in classical theories while mass m and momentum p are used to describe The wavelengths calculated in example 14.5 are negligible compared to the size of the moving objects as well as to the widths of the paths on which the objects are moving.
320 Do you know? Use your brain power The wavelength of the electron in above example is comparable to the size of the hole through which the electron is passing. The wavelength associated with this electron is same as the size of a helium atom and more than double the size of a hydrogen atom. obstacles, or are not measurable, we can use Newtonian mechanics. In conclusion, for both electromagnetic radiation and atomic and sub-atomic particles, particle nature is dominant during their interaction with matter. On the other hand, while traveling through space, particularly when their confinement is of same order of magnitude as their associated wavelength, the wave nature is dominant. Example 14.6: Calculate the de Broglie wavelength of an electron moving with kinetic energy of 100 eV passing through a circular hole of diameter 2 Å. Solution: Given EK = 100 eV = 100 × 1.6 × 10-19 J. The speed of the electron is given by the relation 1 2 mv2 = 100 × 1.6 × 10-19 J. ∴ v = 2 100 1 6 10 9 11 10 19 31 fi fi fi fi ff ff . . J kg = 0.593 × 107 m/s and momentum p = 9.11 × 10-31 kg × 0.593 × 107 m/s = 5.40 × 10-24 kg m/s ∴ the de Broglie wavelength λ = h/p = 6.63 × 10-34 J s / 5.40 × 10-24 kg m/s = 1.23 × 10-10 m = 1.23 Å. Therefore the wavelengths associated with macroscopic particles do not play any significant role in our everyday life and we need not consider their wave nature. Also the wavelengths for macroscopic particles are so small that they cannot be measured. On the other hand, if we try to estimate the associated de Broglie wavelength of a moving electron passing through a small aperture of size 10-10 m or an oxygen molecule in air, we will find it to be significant as can be seen in the following example. On what scale or under which circumstances is the wave nature of matter apparent? Photon picture allows transfer of energy and momentum in the same manner as in Newtonian mechanics. Wave nature does not modify that. Whenever wavelengths are small compared to the dimensions of slits or We have seen earlier that electrons are bound inside a metal surface and need some minimum energy equal to the work function to be knocked off from the surface. This energy, if provided by any means, can make the electron come out of the metal surface. Physical ways to provide this energy differentiate the physical processes involved and accordingly different devices and characterizing microscopes based on them have been designed by scientists. • Thermionic emission : By heating to temperatures ~2000 ºC provide thermal energy. • Field emission : By establishing strong electric fields ~106 V/m at the surface of a metal tip, provide electrical energy. • Photo-electron emission : By shining radiation of suitable frequency (ultraviolet or visible) on a metal surface provide light energy. Electron microscope: You have learnt about resolving power and resolution of telescopes and microscopes that use the ordinary visible light. The resolution of a microscope is limited by the wavelength of the light used. The shorter the wavelength of the characterizing probe, the smaller is the limit of resolution of a microscope, i.e., the resolution of microscope is better. Better
321 Internet my friend resolution can be attained by illuminating the objects to be seen by radiation of smaller wavelengths. We have seen that an electron can behave as a wave and its wavelength is much smaller than the wavelength of visible light. The wavelength can be made much smaller as it depends on the velocity and kinetic energy of the electron. An electron beam accelerated to several keV of energy will correspond to de Broglie wavelength much smaller than an angstrom, i.e., λe << 1×10-10 m. The resolution of this electron microscope will be several hundred times higher than that obtainable with an optical microscope. Other advantages of electron microscopes are that (i) electrons do not penetrate the matter as visible light or X-rays do, (ii) electron beams can be more easily produced and controlled by electric and magnetic fields than electromagnetic waves and (iii) electrons can be focused like light is focused with lenses. It was proposed in 1925 that atoms in the solids can act as diffraction centers for electron waves and can give information about the geometry or structure of solid, just as X-rays do on getting diffracted by solids. However, it took many years to realize an electron microscope for practical applications. The first electron microscope was developed by Herald Ruska in Berlin, Germany in the year 1929. Microscopic objects, when illuminated using electron beams, yield high resolution images. Images of microscopic and nanometric objects and even of viruses have been obtained by scientists using electron microscopes, making valuable contributions to mankind. Transmission electron microscopy can resolve very small particles. A micrograph 1. http://phet-web.colorado.edu/ simulations/schrodinger/dg.jnlp 2. https://physics.info/photoelectric/ 3. https://www.britannica.com/science/ photoelectric-effect 4. https://www.britannica.com/science/ wave-particle-duality 5. https://www.sciencedaily.com/terms/ wave-particle_duality.htm 6. https://www.thoughtco.com/debroglie-hypothesis-2699351 7. https://www.toppr.com/guides/physics/ dual-nature-of-radiation-and-matter 8. http://hyperphysics.phy-astr.gsu.edu/ hbase/quantum/DavGer2.html on the cover page of this book shows tiny crystals of dimensions less than 50 nm. An electron diffraction pattern is also seen on the cover page (spot pattern). When an electron beam passes through a crystal having periodic arrangement of atoms, diffraction occurs. The crystal acts as a collection of diffraction slits for the electron beam.
322 1. Choose the correct answer. i) A photocell is used to automatically switch on the street lights in the evening when the sunlight is low in intensity. Thus it has to work with visible light. The material of the cathode of the photo cell is (A) zinc (B) aluminum (C) nickel (D) potassium ii) Polychromatic (containing many different frequencies) radiation is used in an experiment on photoelectric effect. The stopping potential (A) will depend on the average wavelength (B) will depend on the longest wavelength (C) will depend on the shortest wavelength (D) does not depend on the wavelength iii) An electron, a proton, an α-particle and a hydrogen atom are moving with the same kinetic energy. The associated de Broglie wavelength will be longest for (A) electron (B) proton (C) α-particle (D) hydrogen atom iv) If NRed and NBlue are the number of photons emitted by the respective sources of equal power and equal dimensions in unit time, then (A) NRed < NBlue (B) NRed = NBlue (C) NRed > NBlue (D) NRed ≈ NBlue v) The equation E = pc is valid (A) for all sub-atomic particles (B) is valid for an electron but not for a photon (C) is valid for a photon but not for an electron (D) is valid for both an electron and a photon 2. Answer in brief. i) What is photoelectric effect? ii) Can microwaves be used in the experiment on photoelectric effect? iii) Is it always possible to see photoelectric effect with red light? iv) Using the values of work function given in Table 14.1, tell which metal will require the highest frequency of incident radiation to generate photocurrent. v) What do you understand by the term wave-particle duality? Where does it apply? 3. Explain the inverse linear dependence of stopping potential on the incident wavelength in a photoelectric effect experiment. 4. It is observed in an experiment on photoelectric effect that an increase in the intensity of the incident radiation does not change the maximum kinetic energy of the electrons. Where does the extra energy of the incident radiation go? Is it lost? State your answer with explanatory reasoning. 5. Explain what do you understand by the de Broglie wavelength of an electron. Will an electron at rest have an associated de Broglie wavelength? Justify your answer. 6. State the importance of Davisson and Germer experiment. 7. What will be the energy of each photon in monochromatic light of frequency 5×1014 Hz? [Ans : 3.31×10-19 J = 2.07 eV] 8. Observations from an experiment on photoelectric effect for the stopping potential by varying the incident frequency were plotted. The slope of the linear curve was found to be approximately 4.1×10-15 V s. Given that Exercises
323 the charge of an electron is 1.6 × 10-19 C, find the value of the Planck’s constant h. [Ans : 6.56×10-34 J s] 9. The threshold wavelength of tungsten is 2.76 × 10-5 cm. (a) Explain why no photoelectrons are emitted when the wavelength is more than 2.76 × 10-5 cm. (b) What will be the maximum kinetic energy of electrons ejected in each of the following cases (i) if ultraviolet radiation of wavelength λ = 1.80 × 10-5 cm and (ii) radiation of frequency 4×1015 Hz is made incident on the tungsten surface. [Ans: 2.40 eV, 12.07 eV] 10. Photocurrent recorded in the micro ammeter in an experimental set-up of photoelectric effect vanishes when the retarding potential is more than 0.8 V if the wavelength of incident radiation is 4950 Å. If the source of incident radiation is changed, the stopping potential turns out to be 1.2 V. Find the work function of the cathode material and the wavelength of the second source. [Ans: 1.71 eV, 4270 Å] 11. Radiation of wavelength 4500 Å is incident on a metal having work function 2.0 eV. Due to the presence of a magnetic field B, the most energetic photoelectrons emitted in a direction perpendicular to the field move along a circular path of radius 20 cm. What is the value of the magnetic field B? [Ans. : 1.47 × 10-5 T] 12. Given the following data for incident wavelength and the stopping potential obtained from an experiment on photoelectric effect, estimate the value of Planck’s constant and the work function of the cathode material. What is the threshold frequency and corresponding wavelength? What is the most likely metal used for emitter? Incident wavelength (in Å) 2536 3650 Stopping potential (in V) 1.95 0.5 [Ans: 6.42 × 10-34 J s, 2.80 eV, 6.76 × 1014 Hz, 4440 Å, calcium] 13. Calculate the wavelength associated with an electron, its momentum and speed (a) when it is accelerated through a potential of 54 V, [Ans: 0.167 nm, 39.70 ×10-25 kg m s-1, 4.36 ×106 m s-1] (b) when it is moving with kinetic energy of 150 eV. [Ans: 0.100 nm, 66.13×10-25 kg m s-1, 7.26 ×106 m s-1 ] 14. The de Broglie wavelengths associated with an electron and a proton are same. What will be the ratio of (i) their momenta (ii) their kinetic energies? [Ans: 1, 1836] 15. Two particles have the same de Broglie wavelength and one is moving four times as fast as the other. If the slower particle is an α-particle, what are the possibilities for the other particle? [Ans: proton or neutron] 16. What is the speed of a proton having de Broglie wavelength of 0.08 Å? [Ans: 49.57 × 103 m s-1] 17. In nuclear reactors, neutrons travel with energies of 5 × 10-21 J. Find their speed and wavelength. [Ans: 2.45 × 103 m s-1, 1.62 Å] 18. Find the ratio of the de Broglie wavelengths of an electron and a proton when both are moving with the (a) same speed, (b) same energy and (c) same momentum? State which of the two will have the longer wavelength in each case? [Ans: (a) 1836, (b) electron; 42.85, electron; (c) 1, equal]
324 Can you recall? 15.1. Introduction: Greek philosophers Leucippus (-370 BC) and Democritus (460 – 370 BC) were the first scientists to propose, in the 5th century BC, that matter is made of indivisible parts called atoms. Dalton (1766-1844) gave his atomic theory in early nineteenth century. According to his theory (i) matter is made up of indestructible particles, (ii) atoms of a given element are identical and (iii) atoms can combine with other atoms to form new substances. That atoms were indestructible was shown to be wrong by the experiments of J. J. Thomson (1856-1940) who discovered electrons in 1887. He then proceeded to give his atomic model which had some deficiencies and was later improved upon by Ernest Rutherford (1871- 1937) and Niels Bohr (1885-1962). We will discuss these different models in this Chapter. You have already studied about atoms and nuclei in XIth Std. in chemistry. This chapter will enable you to consolidate your concepts in this subject. We will learn that, an atom contains a tiny nucleus whose size (radius) is about 100000 times smaller than the size of an atom. The nucleus contains all the positive charge of the atom and also 99.9% of its mass. In this Chapter we will also study properties of the nucleus, the forces that keep it intact, its radioactive decays and about the energy that can be obtained from it. 15.2. Thomson’s Atomic Model: Thomson performed several experiments with glass vacuum tube wherein a voltage was applied between two electrodes inside an evacuated tube. The cathode was seen to emit rays which produced a glow when they struck the glass behind the anode. By studying the properties of these rays, he concluded that the rays are made up of negatively charged particles which he called electrons. This demonstrated that atoms are not indestructible. They contain electrons which are emitted by the cathode. Thomson proposed his model of an atom in 1903. According to this model an atom is a sphere having a uniform positive charge in which electrons are embedded. This model is referred to as Plum-pudding model. The total positive charge is equal to the total negative charge of electrons in the atom, rendering it electrically neutral. As the whole solid sphere is uniformly positively charged, the positive charge cannot come out and only the negatively charged electrons which are small, can be emitted. The model also explained the formation of ions and ionic compounds. However, further experiments on structure of atoms which are described below, showed the distribution of charges to be very different than what was proposed in Thomson’s model. 15.3 Geiger-Marsden Experiment: In order to understand the structure of atoms, Rutherford suggested an experiment for scattering of alpha particles by atoms. Alpha particles are helium nuclei and are positively charged (having charge of two protons). The experiment was performed by his colleagues Geiger (1882-1945) and Marsden (1889- 1970) between 1908 and 1913. A sketch of the experimental set up is shown in Fig.15.1. Alpha particles from a source were collimated, i.e., focused into a narrow beam, and were made to fall on a gold foil. The scattered particles produced scintillations on the 15. Structure of Atoms and Nuclei 1. What is Dalton’s atomic model? 2. What are atoms made of? 3. What is wave particle duality? 4. What are matter waves?
325 surrounding screen. The scintillations could be observed through a microscope which could be moved to cover different angles with respect to the incident beam. It was found that most alpha particles passed straight through the foil while a few were deflected (scattered) through various scattering angles. A typical scattering angle is shown by θ in the figure. Only about 0.14% of the incident alpha particles were scattered through angles larger than 0.1o . Even out of these, most were deflected through very small angles. About one alpha particle in 8000 was deflected through angle larger than 90o and a fewer still were deflected through angles as large as 180o . 15.4. Rutherford’s Atomic Model: Results of Geiger-Marsden’s experiment could not be explained by Thomson’s model. In that model, the positive charge was uniformly spread over the large sphere constituting the atom. The volume density of the positive charge would thus be very small and all of the incident alpha particles would get deflected only through very small angles. Rutherford argued that the alpha particles which were deflected back must have encountered a massive particle with large positive charge so that it was repelled back. From the fact that extremely small number of alpha particles turned back while most others passed through almost undeflected, he concluded that the positively charged particle in the atom must be very small in size and must contain most of the mass of the atom. From the experimental data, the size of this particle was found to be about 10 fm (femtometre, 10-15) which is about 10-5 times the size of the atom. The volume of this particle was thus found to be about 10-15 times that of an atom. He called this particle the nucleus of an atom. He proposed that the entire positive charge and most (99.9%) of the mass of an atom is concentrated in the central nucleus and the electrons revolve around it in circular orbits, similar to the revolution of the planets around the Sun in the Solar system. The revolution of the electrons was necessary as without it, the electrons would fall into the positively charged nucleus and the atom would collapse. The space between the orbits of the electrons (which decide the size of the atom) and the nucleus is mostly empty. Thus, most alpha particles pass through this empty space undeflected and a very few which are in direct line with the tiny nucleus or are extremely close to it, get repelled and get deflected through large angles. This model also explains why no positively charged particles are emitted by atoms while negatively charged electrons are. This is because of the large mass of the nucleus which does not get affected when force is applied on the atom. 15.4.1. Difficulties with Rutherford’s Model: Though this model in its basic form is still accepted, it faced certain difficulties. We know from Maxwell’s equations that an accelerated charge emits electromagnetic radiation. An electron in Rutherford’s model moves uniformly along a circular orbit around the nucleus. Even though the magnitude of its velocity is constant, its direction changes continuously and so the motion is an accelerated motion. Thus, the electron should emit electromagnetic radiation continuously. Also, as it emits radiation, its energy would decrease and consequently, the radius of its orbit would decrease continuously. It would then spiral into the nucleus, causing the atom to collapse and lose its atomic properties. As the electron loses energy, its velocity changes continuously and the frequency of the radiation emitted would also change continuously as Fig.15.1: Geiger-Marsden experiment. θ
326 it moves towards the nucleus. None of these things are observed. Firstly, most atoms are very stable and secondly, they do not constantly emit electromagnetic radiation and definitely not of varying frequency. The atoms have to be given energy, e.g., by heating, for them to be able to emit radiation and even then, they emit electromagnetic radiations of particular frequencies as will be seen in the next section. Rutherford’s model failed on all these counts. 15.5 Atomic Spectra: We know that when a metallic object is heated, it emits radiation of different wavelengths. When this radiation is passed through a prism, we get a continuous spectrum. However, the case is different when we heat hydrogen gas inside a glass tube to high temperatures. The emitted radiation has only a few selected wavelengths and when passed through a prism we get what is called a line spectrum as shown for the visible range in Fig.15.2. It shows that hydrogen emits radiations of wavelengths 410, 434, 486 and 656 nm and does not emit any radiation with wavelengths in between these wavelengths. The lines seen in the spectrum are called emission lines. Hydrogen atom also emits radiation at some other values of wavelengths in the ultraviolet (UV), the infrared (IR) and at longer wavelengths. The spectral lines can be divided into groups known as series with names of the scientists who studied them. The series, starting from shorter wavelengths and going to larger wavelengths are called Lyman Fig.15.2: Hydrogen spectrum. series, Balmer series, Paschen series, Brackett series, Pfund series, etc. In each series, the separation between successive lines decreases as we go towards shorter wavelength and they reach a limiting value. Schematic diagrams for the first three series are shown in Fig.15.3. The limiting value of the wavelength for each series is shown by dotted lines in the figure. The observed wavelengths of the emission lines are found to obey the relation. 1 1 1 2 2 fi ff ffl ffi fl R n m --- (15.1) Here λ is the wavelength of a line, R is a constant and n and m are integers. n = 1, 2, 3,…. respectively, for Lyman, Balmer, Paschen…. series, while m takes all integral values greater than n for that series. The wavelength decreases with increase in m. The difference in wavelengths of successive lines in each series (fixed value of n) can be calculated from Eq. (15.1) and shown to decrease with increase in m. Thus, the successive lines in a given series come closer and closer and ultimately reach the values of fi ff n R 2 in the limit m → ∞, for different values of n. Atoms of other elements also emit line spectra. The wavelengths of the lines emitted by each element are unique, so much so that we can identify the element from the wavelengths of the spectral lines that it emits. Rutherford’s model could not explain the atomic spectra. Fig.15.3: Lyman, Balmer and Paschen series in hydrogen spectrum. UV visible IR λ λ λ
327 15.6. Bohr’s Atomic Model: Niels Bohr modified Rutherford’s model by applying ideas of quantum physics which were being developed at that time. He realized that Rutherford’s model is essentially correct and all that it needs is stability of the orbits. Also, the electrons in these stable orbits should not emit electromagnetic waves as required by classical (Maxwell’s) electromagnetic theory. He made three postulates which defined his atomic model. These are given below. 1. The electrons revolve around the nucleus in circular orbits. This is the same assumption as in Rutherford’s model and the centripetal force necessary for the circular motion is provided by the electrostatic force of attraction between the electron and the nucleus. 2. The radius of the orbit of an electron can only take certain fixed values such that the angular momentum of the electron in these orbits is an integral multiple of h/2π, h being the Planck’s constant. Such orbits are called stable orbits or stable states of the electrons and electrons in these orbits do not emit radiation as is demanded by classical physics. Thus, different orbits have different and definite values of angular momentum and therefore, different values of energies. 3. An electron can make a transition from one of its orbit to another orbit having lower energy. In doing so, it emits a photon of energy equal to the difference in its energies in the two orbits. 15.6.1. Radii of the Orbits: Using first two postulates we can study the entire dynamics of the circular motion of the electron, including its energy. Let the mass of the electron be me , its velocity in the nth stable orbit be vn and the radius of its orbit be rn . The angular momentum is then me vn rn and according to the second postulate above, we can write me vn r n h n fi 2ff --- (15.2) The positive integer n is called the principal quantum number of the electron. The centripetal force necessary for the circular motion of the electron is provided by the electrostatic force of attraction between the electron and the nucleus. Assuming the atomic number (number of electrons) of the atom to be Z, the total positive charge on the nucleus is Ze and we can write, m r Ze r e n n n v2 2 0 2 4 fi ffffl --- (15.3) Here, ε 0 is the permeability of vacuum and e is the electron charge. Eliminating vn from the Eq.(15.2) and Eq.(15.3), we get, m n h m r e e n 2 2 2 2 3 4π = Ze rn 2 0 2 4fiff ∴ r n h m Ze n e fi fi fi 2 2 0 2 ff ffl --- (15.4) Similarly, eliminating rn from Eq.(15.2) and Eq.(15.3), we get, vn fi Ze h n 2 0 2fflff --- (15.5) Equation (15.4) shows that the radius of the orbit is proportional to n 2 , i.e., the square of the principal quantum number. The radius increases with increase in n. The hydrogen atom has only one electron, i.e., Z is 1. Substituting the values of the constants h, ε0 , m and e in Eq.(15.4), we get, for n = 1, r1 = 0.053 nm. This is called the Bohr radius and is denoted by a0 = h m ee 2 0 2 fi ff . This is the radius of the smallest orbit of the electron in hydrogen atom. From Eq. (15.4), we can write, r a n n = 0 2 --- (15.6) Example 15.1: Calculate the radius of the 3rd orbit of the electron in hydrogen atom. Solution: The radius of nth orbit is given by r a n n = 0 2 . Thus, the radius of the third orbit (n = 3) is r a a 3 0 2 0 fi fi 3 9 fi ff9 0 .053nm = 0.477 nm.
328 15.6.2. Energy of the Electrons: The total energy of an orbiting electron is the sum of its kinetic energy and its electrostatic potential energy. Thus, E K n n fi ff . . E Pff .E E ,ffffffff being the total energy of an electron in the nth orbit. E m Ze r n e n n fi ffffl ffi fl 1 2 4 2 2 0 v . Using Eq. (15.3) and (15.4) this gives E m Z e h n n e fi ff 2 4 0 2 2 8ffl --- (15.7) The negative value of the energy of the electron indicates that the electron is bound inside the atom and it has to be given energy so as to make the total energy zero, i.e., to make the electron free from the nucleus. The energy increases (becomes less negative) with increase in n. Substituting the values of the constants m e,ff ,ffhff ff and ε 0 in the above equation, we get E Z n n fi fffi . 13 6 2 2 eV --- (15.8) The first orbit ( n = 1) which has minimum energy, is called the ground state of the atom. Orbits with higher values of n and therefore, higher values of energy are called the excited states of the atom. If the electron is in the nth orbit, it is said to be in the nth energy state. For hydrogen atom (Z = 1) the energy of the electron in its ground state is -13.6 eV and the energies of the excited states increase as given by Eq.(15.8). The energy levels of hydrogen atom are shown in Fig.15.4. The energies of the levels are given in eV. Fig.15.4: Energy levels and transitions between them for hydrogen atom (energy not to scale). The energy levels come closer and closer as n increases and their energy reaches a limiting value of zero as n goes to infinity. The energy required to take an electron from the ground state to an excited state is called the excitation energy of the electron in that state. For hydrogen atom, the minimum excitation energy (of n = 2 state) is -3.4-(-13.6) =10.2 eV. Example 15.2: In a Rutherford scattering experiment, assume that an incident alpha particle (radius 1.80 fm) is moving directly toward a target gold nucleus (radius 6.23 fm). If the alpha particle stops right at the surface of the gold nucleus, how much energy did it have to start with? Solution: Initially when the alpha particle is far away from the gold nucleus, its total energy is equal to its kinetic energy. As it comes closer to the nucleus, more and more of its kinetic energy gets converted to potential energy. By the time it reaches the surface of the nucleus, its kinetic energy is completely converted into potential energy and it stops moving. Thus, the initial kinetic energy K, of the alpha particle is equal to the potential energy when it is at the surface of the nucleus, i.e., when the distance between the gold nucleus and the alpha particle is equal to the radius of the gold nucleus. ∴ K e Ze r r fi ffl ffi ff fi fi 1 fi fi 4 2 fl 0 1 2 , where, Z is the atomic number of gold and r r 1 2 ffl ffl and are the radii of the gold nucleus and alpha particle respectively. For gold Z = 79. fi ff ffi fl ffl ff ffi fl ffl K Ze r r fi fi fi fi . . 1 4 2 9 10 2 79 1 6 10 6 23 1 0 2 1 2 9 19 2 . . fi . fiMeV 80 10 4 53 10 28 31 15 12 ffi fl ff ff J
329 In order to remove or take out the electron in the ground state from a hydrogen atom, i.e., to make it free (and have zero energy), we have to supply 13.6 eV energy to it. This energy is called the ionization energy of the hydrogen atom. The ionization energy of an atom is the minimum amount of energy required to be given to an electron in the ground state of that atom to set the electron free. It is the binding energy of hydrogen atom. If we form a hydrogen atom by bringing a proton and an electron from infinity and combine them, 13.6 eV energy will be released. According to the third postulate of Bohr, when an electron makes a transition from mth to nth orbit (m > n), the excess energy E E m n − is emitted in the form of a photon. The energy of the photon which can be written as hv, v being its frequency, is therefore given by, h m Z e h n m e fi ff ffl ffi fl 2 4 0 2 2 2 8 1 1 which can be written in terms of the wavelength as 1 8 1 1 2 4 0 3 2 2 fi ff ffl ffi fl m Z e c h n m e ---(15.9) Here c is the velocity of light in vacuum. We define a constant called the Rydberg’s constant, RH as RH = m e c h e 4 0 3 8 ε = 1.097 × 107 m–1. ---(15.10) In terms of R, the wavelength is given by 1 1 2 1 2 2 fi ff ffl ffi fl R Z n m H ---(15.11) This is called the Rydberg’s formula. Remember that for hydrogen Z is 1. Thus, Eq.(15.11) correctly describes the observed spectrum of hydrogen as given by Eq.(15.1). Example 15.3: Determine the energies of the first two excited states of the electron in hydrogen atom. What are the excitation energies of the electrons in these orbits? Solution: The energy of the electron in the nth orbit is given by E n n fi fffi . 13 6 1 2 eV. The first two excited states have n = 2 and 3. Their energies are E2 2 13 6 1 2 fiff fiff . . 3 4 eV and E3 2 13 6 1 3 fiff fiff . . 1 51 eV . Excitation energy of an electron in nth orbit is the difference between its energy in that orbit and the energy of the electron in its ground state, i.e. -13.6 eV. Thus, the excitation energies of the electrons in the first two excited states are 10.2 eV and 12.09 eV respectively. Example 15.4: Calculate the wavelengths of the first three lines in Paschen series of hydrogen atom. Solution: The wavelengths of lines in Paschen series (n=3) are given by 1 1 1 1 097 10 1 3 1 2 2 7 2 2 fi ff ffl ffi fl ff ffl ffi fl R . n m m H m-1 for m = 4,5,…. For the first three lines in the series, m = 4, 5 and 6. Substituting in the above formula we get, 1 1 097 10 1 3 1 4 1 097 10 7 9 16 1 7 2 2 7 fi ff ffl ffi fl ff ffl ffl ffl . . / ( ) ff ffl mffi 0 0533 107 1 . fi1 ff 1 876 . x 10-6 m 1 1 097 10 1 3 1 5 1 097 10 16 9 25 2 7 2 2 7 fi ff ffl ffi fl ff ffl ffl ffl . . / ( ) m m ff ff ffl ffl ffi ffi 0 075 1 282 10 10 7 2 6 1 . fi . 1 1 097 10 1 3 1 6 1 097 10 27 9 36 3 7 2 2 7 fi ff ffl ffi fl ff ffl ffl ffl . . / ( ) m m ff ff ffl ffl ffi ffi 0 0914 1 094 10 10 7 3 6 1 . fi .
330 15.6.3. Limitations of Bohr’s Model: Even though Bohr’s model seemed to explain hydrogen spectrum, it had a few shortcomings which are listed below. (i) It could not explain the line spectra of atoms other than hydrogen. Even for hydrogen, more accurate study of the observed spectra showed multiple components in some lines which could not be explained on the basis of this model. (ii) The intensities of the emission lines seemed to differ from line to line and Bohr’s model had no explanation for that. (iii) On theoretical side also the model was not entirely satisfactory as it arbitrarily assumed orbits following a particular condition to be stable. There was no theoretical basis for that assumption. A full quantum mechanical study is required for the complete understanding of the structure of atoms which is beyond the scope of this book. Some reasoning for the third shortcoming (i.e., theoretical basis for the second postulate in Bohr’s atomic model) was given by de Broglie which we consider next. 15.6.4 De Broglie’s Explanation: We have seen in Chapter 14 that material particles also have dual nature like that for light and there is a wave associated with every material particle. De Broglie suggested that instead of considering the orbiting electrons inside atoms as particles, we should view them as standing waves. Similar to the case of standing waves on strings or in pipes as studied in Chapter 6, the length of the orbit of an electron has to be an integral multiple of its wavelength. Thus, the length of the first orbit will be equal to one de Broglie wavelength, λ1 of the electron in that orbit, that of the second orbit will be twice the de Broglie wavelength of the electron in that orbit and so on. This is shown for the 4th orbit in Fig.(15.5) In general, we can write, Fig. 15.5: Standing electron wave for the 4th orbit of an electron in an atom. 2fi ff r n n n ffl , n = 1,2,3….., giving fi ff n nr n ffl 2 --- (15.12) The de Broglie wavelength is related to the linear momentum pn ,ffof the particle by fin n n h p h m ff ff v . Substituting this in Eq. (15.12) gives, p hn r n n fi 2ff . Thus, the angular momentum of the electron in nth orbit, Ln , can be written as L p r n h n n fi fin 2ff , which is the second postulate of Bohr’s atomic model. Therefore, considering electrons as waves gives some theoretical basis for the second postulate made by Bohr. 15.7. Atomic Nucleus: 15.7.1 Constituents of a Nucleus: The atomic nucleus is made up of subatomic, meaning smaller than an atom, particles called protons and neutrons. Together, protons and neutrons are referred to as nucleons. Mass of a proton is about 1836 times that of an electron. Mass of a neutron is nearly same as that of a proton but is slightly higher. The proton is a positively charged particle. The magnitude of its charge is equal to the magnitude of the charge of an electron. The neutron, as the name suggests, is electrically neutral. The number of protons in an atom is called its atomic number and is designated by Z. The number of electrons
331 in an atom is also equal to Z. Thus, the total positive and total negative charges in an atom are equal in magnitude and the atom as a whole is electrically neutral. The number of neutrons in a nucleus is written as N. The total number of nucleons in a nucleus is called the mass number of the atom and is designated by A = Z + N. The mass number determines the mass of a nucleus and of the atom. The atoms of an element X are represented by the symbol for the element and its atomic and mass numbers as Z A X . For example, symbols for hydrogen, carbon and oxygen atoms are written as 1 1 H , 6 12C and 8 16O . The chemical properties of an atom are decided by the number of electrons present in it, i.e., by Z. The number of protons and electrons in the atoms of a given element are fixed. For example, hydrogen atom has one proton and one electron, carbon atom has six protons and six electrons. The number of neutrons in the atoms of a given element can vary. For example, hydrogen nucleus can have zero, one or two neutrons. These varieties of hydrogen are referred to as 1 1 H , 1 2 H and 1 3 H and are respectively called hydrogen, deuterium and tritium. Atoms having the same number of protons but different number of neutrons are called iosotopes. Thus, deuterium and tritium are isotopes of hydrogen. They have the same chemical properties as those of hydrogen. Similarly, helium nucleus can have one or two neutrons and are referred as 2 3 He and 2 4 He . The atoms having the same mass number A, are called isobars. Thus, 1 3 H and 2 3 He are isobars. Atoms having the same number of neutrons but different values of atomic number Z, are called iosotones. Thus, 1 3 H and 2 4 He are isotones. Units for measuring masses of atoms and subatomic particles Masses of atoms and subatomic particles are measured in three different units. First unit is the usual unit kg. The masses of electron, proton and neutron, me , m p and mn respectively, in this unit are: me = 9.109383 × 10-31 kg m p = 1.672623 × 10-27 kg mn = 1.674927 × 10-27 kg Obviously, kg is not a convenient unit to measure masses of atoms or subatomic particles which are extremely small compared to one kg. Therefore, another unit called the unified atomic mass unit (u) is used for the purpose. One u is equal to 1/12th of the mass of a neutral carbon atom having atomic number 12, in its lowest electronic state. 1 u = 1.6605402 × 10-27 kg. In this unit, the masses of the above three particles are me = 0.00055 u m p = 1.007825 u mn = 1.008665 u. The third unit for measuring masses of atoms and subatomic particles is in terms of the amount of energy that their masses are equivalent to. According to Einstein’s famous mass-energy relation, a particle having mass m is equivalent to an amount of energy E = mc2 . The unit used to measure masses in terms of their energy equivalent is the eV/c2 . One atomic mass unit is equal to 931.5 MeV/ c2. The masses of the three particles in this unit are me = 0.511 MeV/c2 m p = 938.28 MeV/c2 mn = 939.57 MeV/c2 15.7.2. Sizes of Nuclei: The size of an atom is decided by the sizes of the orbits of the electrons in the atom. Larger the number of electrons in an atom, higher are the orbits occupied by them and larger is the size of the atom. Similarly, all nuclei do not have the same size. Obviously, the size of a nucleus depends on the number of nucleons present in it, i.e., on its atomic number A. From experimental observations it has been found that the radius RX of a nucleus X is related to A as R R X = 0A 1 3 --- (15.13)
332 where R0 = 1.2 x 10-15 m. The density ρ inside a nucleus is given by 4 3 3 fi ff R mffl A,ffiwhere, we have assumed m to be the average mass of a nucleon (proton and neutron) as the difference in their masses is rather small. The density is then given by, fi ff ffl 3 4 3 mA RX Substituting for RX from Eq.(15.13), we get, fi ff ffl ffl 3 4 0 3 m R constant. Thus, the density of a nucleus does not depend on the atomic number of the nucleus and is the same for all nuclei. Substituting the values of the constants m, π and R0 the value of the density is obtained as 2.3 x 1017 kg m-3 which is extremely large. Among all known elements, osmium is known to have the highest density which is only 2.2 x 104 kg m-3. This is smaller than the nuclear density by thirteen orders of magnitude. distance up to which it is effective. Over short distances of about a few fm, the strength of the nuclear force is much higher than that of the other two forces. Its range is very small and its strength goes to zero when two nucleons are at a distance larger than a few fm. This is in contrast to the ranges of electrostatic and gravitational forces which are infinite. The protons in the nucleus repel one another due to their similar (positive) charges. The nuclear forces between the nucleons counter the forces of electrostatic repulsion. As nuclear force is much stronger than the electrostatic force for the distances between nucleons in a typical nucleus, it overcomes the repulsive force and keeps the nucleons together, making the nucleus stable. The nuclear force is not yet well understood. What we know about its properties can be summarized as follows. 1. It is the strongest force among subatomic particles. Its strength is about 50-60 times larger than that of the electrostatic force. 2. Unlike the electromagnetic and gravitational forces which act over large distances (their range is infinity), the nuclear force has a range of about a few fm and the force is negligible when two nucleons are separated by larger distances. 3. The nuclear force is independent of the charge of the nucleons, i.e., the nuclear force between two neutrons with a given separation is the same as that between two protons or between a neutron and a proton at the same separation. 15.8. Nuclear Binding Energy: We have seen that in a hydrogen atom, the energy with which the electron in its ground state is bound to the nucleus (which is a single proton in this case) is 13.6 eV. This is the amount of energy which is released when a proton and an electron are brought from infinity to form the atom in its ground state. In other words, this is the amount of energy which Example 15.5: Calculate the radius and density of 70Ge nucleus, given its mass to be approximately 69.924 u. Solution: The radius of a nucleus X with mass number A is given by R R X = 0A 1 3 , where R0 = 1.2 × 10-15 m Thus, the radius of 70Ge is RGe = 1.2 × 10-15 × 701/3 = 4.95 × 10-15 m. The density is given by fi ff ffl 3 4 0 3 m R . ∴ fi ff ffl ffi ffi ffi fl ffi 3 69 924 1 66 10 4 4 95 10 27 15 3 . . . = 2.29 × 1017 kg m-3. 15.7.3 Nuclear Forces: You have learnt about the four fundamental forces that occur in nature. Out of these four, the force that determines the structure of the nucleus is the strong force, also called the nuclear force. This acts between protons and neutrons and is mostly attractive. It is different from the electrostatic and gravitational force in terms of its strength and range, i.e., the
333 has to be supplied to the atom to separate the electron and the proton, i.e., to make them free. The protons and the neutrons inside a nucleus are also bound to one another. Energy has to be supplied to the nucleus to make the nucleons free, i.e., separate them and take them to large distances from one another. This energy is the binding energy of the nucleus. Same amount of energy is released if we bring individual nucleons from infinity to form the nucleus. Where does this released energy come from? It comes from the masses of the nucleons. The mass of a nucleus is smaller than the total mass of its constituent nucleons. Let the mass of a nucleus having atomic number Z and mass number A be M. It is smaller than the sum of masses of Z protons and N (= A-Z) neutrons. We can write, fiM Z ff ffl fi fi m N p n m Mffi ---(15.14) ∆M is called the mass defect of the nucleus. The binding energy EB , of the nucleus is given by E M c B fi ff 2 =( ff Z m N mff ) M c p n fi ff 2 ---(15.15) On the right hand side of Eq.(15.15), we can add and subtract the mass of Z electrons which will enable us to use atomic masses in the calculation of binding energy. The Eq.(15.15) thus becomes EB = Zm Zm fi fi N mfi fi M Z m c flff ffl p e fi fi ffin e ff ffl fi 2 E Z m N m M c Z A B H fi ff ffi n ffl fl fl fl 2 ---(15.16) Here, mH is the mass of a hydrogen atom and Z A M is the atomic mass of the element being considered. We will be using atomic masses in what follows, unless otherwise specified. An important quantity in this regard is the binding energy per nucleon (=EB/A) of a nucleus. This can be considered to be the average energy which has to be supplied to a nucleon to remove it from the nucleus and make it free. This quantity thus, allows us to compare the relative strengths with which nucleons are bound in a nucleus for different Deuterium nucleus has the minimum value of EB/A and is therefore, the least stable nucleus. The value of EB/A increases with increase in atomic number and reaches a plateau for A between 50 to 80. Thus, the nuclei of these elements are the most stable among all the species. The peak occurs around A = 56 corresponding to iron, which is thus one of the most stable nuclei. The value of EB/A decreases gradually for values of A greater than 80, making the nuclei of those elements slightly less stable. Note that the binding energy of hydrogen nucleus having a single proton is zero. Fig.15.6: Binding energy per nucleon as a function of mass number. species and therefore, compare their stabilities. Nuclei with higher values of EB/A are more stable as compared to nuclei having smaller values of this quantity. Binding energy per nucleon for different values of A (i.e., for nuclei of different elements) are plotted in Fig.15.6. Example 15.6: Calculate the binding energy of 3 7 Li , the masses of hydrogen and lithium atoms being 1.007825 u and 7.016 u respectively. Solution: The binding energy is given by EB = ( ff 3 4ff ) 2 m m m c H n fi ff Li = (3 × 1.007825 + 4 × 1.00866 - 7.016) × 931.5 = 39.23 MeV 15.9 Radioactive Decays: Many of the nuclei are stable, i.e., they can remain unchanged for a very long time. These have a particular ratio of the mass number and the atomic number. Other nuclei occurring in nature, are not so stable and undergo changes
334 Do you know? in their structure by emission of some particles. They change or decay to other nuclei (with different A and Z) in the process. The decaying nucleus is called the parent nucleus while the nucleus produced after the decay is called the daughter nucleus. The process is called radioactive decay or radioactivity and was discovered by Becquerel (1852-1908) in 1876. Radioactive decays occur because the parent nuclei are unstable and get converted to more stable daughter nuclei by the emission of some particles. These decays are of three types as described below. Alpha Decay: In this type of decay, the parent nucleus emits an alpha particle which is the nucleus of helium atom. The parent nucleus thus loses two protons and two neutrons. The decay can be expressed as Z A Z A X fi ff ffl Y ffl 2 4 ffi --- (15.17) X is the parent nucleus and Y is the daughter nucleus. All nuclei with A > 210 undergo alpha decay. The reason is that these nuclei have a large number of protons. The electrostatic repulsion between them is very large and the attractive nuclear forces between the nucleons are not able to cope with it. This makes the nucleus unstable and it tries to reduce the number of its protons by ejecting them in the form of alpha particles. An example of this is the alpha decay of bismuth which is the parent nucleus with A = 212 and Z = 83. The daughter nucleus has A= 208 and Z = 81, which is thallium. The reaction is 83 212 81 208 Bi fi ff Tl ffl The total mass of the products of an alpha decay is always less than the mass of the parent atom. The excess mass appears as the kinetic energy of the products. The difference in the energy equivalent of the mass of the parent atom and that of the sum of masses of the products is called the Q-value, Q, of the decay and is equal to the kinetic energy of the products. We can write, Q = [mX – mY – mHe]c2 , --- (15.18) Beta Decay: In this type of decay the nucleus emits an electron produced by converting a neutron in the nucleus into a proton. Thus, the basic process which takes place inside the parent nucleus is n → p + e- + antineutrino --- (15.19) Neutrino and antineutrino are particles which Becquerel discovered the radioactive decay by chance. He was studying the X-rays emitted by naturally occurring materials when exposed to Sunlight. He kept a photographic plate covered in black paper, separated from the material by a silver foil. When the plates were developed, he found images of the material on them, showing that the X-rays could penetrate the black paper and silver foil. Once while studying uranium-potassium phosphate in a similar way, the Sun was behind the clouds so no exposure to Sunlight was possible. In spite of this, he went ahead and developed the plates and found images to have formed. With further experimentation he concluded that some rays were emitted by uranium itself for which no exposure to Sunlight was necessary. He then passed the rays through magnetic field and found that the rays were affected by the magnetic field. He concluded that the rays must be charged particles and hence were different from the X-rays. The term radioactivity was coined by Marie Curie who made further studies and later discovered element radium along with her husband. The Nobel Prize for the year 1903 was awarded jointly to Becquerel, Marie Curie and Pierre Curie for their contributions to radioactivity. mX, mY and mHe being the masses of the parent atom, the daughter atom and the helium atom. Note that we have used atomic masses to calculate the Q factor.
335 have very little mass and no charge. During beta decay, the number of nucleons i.e., the mass number of the nucleus remains unchanged. The daughter nucleus has one less neutron and one extra proton. Thus, Z increases by one and N decreases by one, A remaining constant. The decay can be written as, Z A Z A X Y fi ff ff e ffffl 1 antineutrino --- (15.20) An example is 27 60 28 60 Co fi ff Ni e ffffl antineutrino. There is another type of beta decay called the beta plus decay in which a proton gets converted to a neutron by emitting a positron and a neutrino. A positron is a particle with the same properties as an electron except that its charge is positive. It is known as the antiparticle of electron. This decay can be written as, p → n + e+ + neutrino --- (15.21) The mass number remains unchanged during the decay but Z decreases by one and N increases by one. The decay can be written as Z A Z A X fi ff ffl Y e ffff 1 neutrino --- (15.22) An example is 11 22 10 22 Na fi ff Ne e ffff neutrino. An interesting thing about beta plus decay is that the mass of a neutron is higher than the mass of a proton. Thus the decay described by Eq. (15.21) cannot take place for a free proton. However, it can take place when the proton is inside the nucleus as the extra energy needed to produce a neutron can be obtained from the rest of the nucleus. In beta decay also, the total mass of the products of the decay is less than the mass of the parent atom. The excess mass is converted into kinetic energy of the products. The Q value for the decay can be written as Q = [mX – mY – me ]c2 --- (15.23) Here, we have ignored the mass of the neutrino as it is negligible compared to the masses of the nuclei. Example 15.7: Calculate the energy released in the alpha decay of 238Pu to 234U, the masses involved being mPu = 238.04955 u, mU = 234.04095 u and mHe = 4.002603 u. Solution: The decay can be written as 238Pu → 234U + 4 He. Its Q value, i.e., the energy released is given by Q = [mPu -mU – mHe]c2 = [238.04955 - 234.04095 - 4.002603]c2 u = 0.005997 × 931.5 MeV = 5.5862 MeV. Example 15.8: Calculate the maximum kinetic energy of the beta particle (positron) emitted in the decay of 11 22 Na , given the mass of 11 22 Na = 21.994437 u, 10 22 Ne = 21.991385 u and me = 0.00055 u. Solution: The decay can be written as 11 22 10 22 Na fi ff Ne e ffff neutrino. The energy released is Q = [mNa -mNe -me ]c2 = [21.994437-21.991385-0.00055]c2 = 0.002502 × 931.5 MeV = 2.3306 MeV This is the maximum energy that the beta particle (e+ ) can have, the neutrino having zero energy in that case. Gamma Decay: In this type of decay, gamma rays are emitted by the parent nucleus. As you know, gamma ray is a high energy photon. The daughter nucleus is same as the parent nucleus as no other particle is emitted, but it has less energy as some energy goes out in the form of the emitted gamma ray. We have seen that the electrons in an atom are arranged in different energy levels (orbits) and an electron from a higher orbit can make a transition to the lower orbit emitting a photon in the process. The situation in a nucleus is similar. The nucleons occupy energy levels with different energies. A nucleon can make a transition from a higher energy level to a lower energy level, emitting a photon in the process. The difference between atomic and nuclear energy levels is in their energies and energy separations. Energies and the differences in the
336 Use your brain power energies of different levels in an atom are of the order of a few eVs, while those in the case of a nucleus are of the order of a few keV to a few MeV. Therefore, whereas the radiations emitted by atoms are in the ultraviolet to radio region, the radiations emitted by nuclei are in the range of gamma rays. Usually, the nucleons in a nucleus are in the lowest possible energy state. They cannot easily get excited as a large amount of energy (in keVs or MeVs) is required for their excitation. A nucleon however may end up in an excited state as a result of the parent nucleus undergoing alpha or beta decay. Thus, gamma decays usually occur after one of these decays. For example, 57Co undergoes beta plus decay to form the daughter nucleus 57Fe which is in an excited state having energy of 136 keV. There are two ways in which it can make a transition to its ground state. One is by emitting a gamma ray of energy 136 keV and the other is by emitting a gamma ray of energy 122 keV and going to an intermediate state first and then emitting a photon of energy 14 keV to reach the ground state. Both these emissions have been observed experimentally. Which type of decay a nucleus will undergo depends on which of the resulting daughter nucleus is more stable. Often, the daughter nucleus is also not stable and it undergoes further decay. A chain of decays may take place until the final daughter nucleus is stable. An example of such a series decay is that of 238U, which undergoes a series of alpha and beta decays, a total of 14 times, to finally reach a stable daughter nucleus of 206Pb. that if we have one atom of the radioactive material, we can never predict how long it will take to decay. If we have N0 number of radioactive atoms (parent atoms or nuclei) of a particular kind say uranium, at time t = 0, all we can say is that their number will decrease with time as some nuclei (we cannot say which ones) will decay. Let us assume that at time t, number of parent nuclei which are left is N(t). How many of these will decay in the interval between t and t +dt ? We can guess that the larger the value of N(t), larger will be the number of decays dN in time dt. Thus, we can say that dN will be proportional to N(t). Also, we can guess that the larger the interval dt, larger will be the number of particles decaying in that interval. Thus, we can write, dN fi fffi fi N tffl ffi dt, or, dN fi ff fl N tffl ffi dt --- (15.24) where, λ is a constant of proportionality called the decay constant. The negative sign in Eq.(15.24) indicates that the change in the number of parent nuclei dN, is negative, i.e., N(t) is decreasing with time. We can integrate this equation as N N t t dN N t dt 0 0 fi ff ffl ffl fi ff ffi fl , Here, N0 is the number of parent atoms at time t = 0. Integration gives, ln ffl,fflffl N t N t fi ff ffl ffi 0 fl or, N t N e t fi ff ffl ffi 0 fl ---(15.25) This is the decay law of radioactivity. The rate of decay, i.e., the number of decays per unit time fi dN tff ffl dt , also called the activity A(t), can be written using Eq.(15.24) and (15.25) as, A t dN dt N t N e t fi ff ffl ffi ffl fi ff ffl ffi fl fl fl 0 --- (15.26) At t = 0, the activity is given by A N 0 0 fi ff .ffl Using this, Eq.(15.26) can be written as A t A e t fi ff ffl ffi 0 fl ffff --- (15.27) Why don’t heavy nuclei decay by emitting a single proton or a single neutron? 15.10. Law of Radioactive Decay: Materials which undergo alpha, beta or gamma decays are called radioactive materials. The nature of radioactivity is such
337 Activity is measured in units of becquerel (Bq) in SI units. One becquerel is equal to one decay per second. Another unit to measure activity is curie (Ci). One curie is 3.7 x 1010 decays per second. Thus, 1 Ci = 3.7 x 1010 Bq. 15.10.1. Half-life of Radioactive Material: The time taken for the number of parent radioactive nuclei of a particular species to reduce to half its value is called the half-life T1/2, of the species. This can be obtained from Eq. (15.25) N N e 0 T 0 2 1 2 fi ffffl / , giving e fiT1 2 2 / ff , or T1 2 2 0 693 / ln fi fi . / ff ff --- (15.28) The interesting thing about half-life is that even though the number goes down from N0 to N0 2 in time T1/2 , after another time interval T1/2, the number of parent nuclei will not go to zero. It will go to half of the value at t =T1/2, i.e., to N0 4 . Thus, in a time interval equal to half-life, the number of parent nuclei reduces by a factor of ½. 15.10.2 Average Life of a Radioactive Species: We have seen that different nuclei of a given radioactive species decay at different times, i.e., they have different life times. We can calculate the average life time of a nucleus of the material using Eq.(15.25) as described below. The number of nuclei decaying between time t and t + dt is given by fi fi N e dt t 0 ff . The life time of these nuclei is t. Thus, the average lifetime τ of a nucleus is fi ff ff ff ff ffl ffl ffi fl ffi fl 1 0 0 0 N 0 t N e dt t e dt t t ffl ffl ffl ffl ffl ffl, Integrating the above we get fi ff ffl 1/ --- (15.29) The relation between the average life and halflife can be obtained using Eq.(15.28) as T1 2 2 0 693 / fi fi ff ff ln . --- (15.30) Example 15.9: The half-life of a nuclear species NX is 3.2 days. Calculate its (i) decay constant, (ii) average life and (iii) the activity of its sample of mass 1.5 mg. Solution: The half-life (T1/2) is related to the decay constant (λ) by T1 2 0 693 / fi . / ff giving, fi ff 0 693 . / T1/2 = 0.693/3.2 = 0.2166 per day = 0.2166 /(24 × 3600) s-1 = 2.5 × 10-6 s-1 . Average life is related to decay constant by fi ff ffl 1/ = 1/0.2166 per day = 4.617 days The activity is given by A = λN(t), where N(t) is the number of nuclei in the given sample. This is given by N(t)= 6.02 × 1023 × 1.5 × 10-3/Y = 9.03 × 1020/Y Here, Y is the atomic mass of nuclear species X in g per mol. ∴ A = 9.03 × 1020 × 2.5 × 10-6/Y = 2.257 × 1015 /Y = 2.257 × 1015/(Y x 3.7 × 1010 ) Ci = 6.08 × 104 /Y Ci. Example 15.10: The activity of a radioactive sample decreased from 350 s-1 to 175 s-1 in one hour. Determine the halflife of the species. Solution: The time dependence of activity is given by A t A e t fi ff ffl ffi 0 fl , where, A(t) and A0 are the activities at time t and 0 respectively. 175 = 350 efiff3600 , or, 3600 λ = ln (350/175) = ln 2 = 0.6931 λ = 0.6931/3600 = 1.925 × 10-4 s-1 . The half-life is given by T1 2 0 693 / fi . / ff . ∴ T s 1 2 4 0 693 3 1 925 10 3 6 10 / . . fi . ff fi ff ffl Example 15.11: In an alpha decay, the daughter nucleus produced is itself unstable and undergoes further decay. If the number of parent and daughter nuclei at time t are N p and Nd respectively and their decay
338 15.11. Nuclear Energy: You are familiar with the naturally occurring, conventional sources of energy. These include the fossil fuels, i.e., coal, petroleum, natural gas, and fire wood. The energy generation from these fuels is through chemical reactions. It takes millions of years for these fuels to form. Naturally, the supply of these conventional sources is limited and with indiscriminate use, they are bound to get over in a couple of hundred years from now. Therefore, we have to use alternative sources of energy. The ones already in use are hydroelectric power, solar energy, wind energy and nuclear energy, nuclear energy being the largest source among these. Nuclear energy is the energy released when nuclei undergo a nuclear reaction, i.e., when one nucleus or a pair of nuclei, due to their interaction, undergo a change in their structure resulting in new nuclei and generating energy in the process. While the energy generated in chemical reactions is of the order of few eV per reaction, the amount of energy released in a nuclear reaction is of the order of a few MeV. Thus, for the same weight of fuel, the nuclear energy released is about a million times that released through chemical reactions. However, nuclear energy generation is a very complex and expensive process and it can also be extremely harmful. Let us learn more about it. We have seen in section 15.8 that the mass of a nucleus is smaller than the sum of masses of its constituents. The difference in these two masses is the binding energy of the nucleus. It would be the energy released if the nucleus is formed by bringing together its constituents from infinity. This energy is large (in MeV), and this process can be a good source of energy. In practice, we never form nuclei starting from individual nucleons. However, we can obtain nuclear energy by two other processes (i) nuclear fission in which a heavy nucleus is broken into two nuclei of smaller masses and (ii) nuclear fusion in which two light nuclei undergo nuclear reaction and fuse together to form a heavier nucleus. Both fission and fusion are nuclear reactions. Let us understand how nuclear energy is released in the two processes. 15.11.1. Nuclear Fission: We have seen in Fig.15.6 that the binding energy per nucleon (EB/A) depends on the mass number of the nuclei. This quantity is a measure of the stability of the nucleus. As seen from the figure, the middle weight nuclei (mass number ranging from 50 to 80) have highest binding energy per nucleon and are most stable, while nuclei with higher and lower atomic masses have smaller values of EB/A. The value of EB/A goes on decreasing till A~238 which is the mass number of the heaviest naturally occurring element which is uranium. Many of the heavy nuclei are unstable and decay into two smaller mass nuclei. Let us consider a case when a heavy nucleus, say with A ~230, breaks into two nuclei having A between 50 and 150. The EB/A of the product nuclei will be higher than that constants are λ p and λd respectively. What condition needs to be satisfied in order for Nd to remain constant? Solution: The number of parent nuclei decaying between time 0 and dt, for small values of dt is given by N dt p p λ . This is the number of daughter nuclei produced in time dt. The number of daughter nuclei decaying in the same interval is N dt d d λ . For the number of daughter nuclei to remain constant, these two quantities, i.e., the number of daughter nuclei produced in time dt and the number decaying in time dt have to be equal. Thus, the required condition is given by N dt N dt p p fi fi ff d d , or, N N p p fi fi ff d d
339 of the parent nucleus. This means that the combined masses of the two product nuclei will be smaller than the mass of the parent nucleus. The difference in the mass of the parent nucleus and that of the product nuclei taken together will be released in the form of energy in the process. This process in which a heavy nucleus breaks into two lighter nuclei with the release of energy is called nuclear fission and is a source of nuclear energy. One of the nuclei used in nuclear energy generation by fusion is 92 U 236 . This has a halflife of 2.3 x 107 years and an activity of 6.5 x 10-5 Ci/g. However, it being fissionable, most of its nuclei have already decayed and it is not found in nature. More than 99% of natural uranium is in the form of 92 U 238 and less than 1% is in the form of 92 U 235 . 92 U 238 also decays, but its half-life is about 103 times higher than that of 92 U 236 and is therefore not very useful for energy generation. The species needed for nuclear energy generation, i.e., 92 U 236 can be obtained from the naturally occurring 92 U 235 by bombarding it with slow neutrons. 92 U 235 absorbs a neutron and yields 92 U 236 . This reaction can be written as 92 U 235 + n→ 92 U 236 . 92 U 236 can undergo fission in several ways producing different pairs of daughter nuclei and generating different amounts of energy in the process. Some of its decays are 92 U 236 → I + Y + 2n 53 137 39 97 92 U 236 → Ba + Kr + 2n 56 140 36 94 92 U 236 → Sb + Nb + 4n 51 133 41 99 --- (15.31) Some of the daughter nuclei produced are not stable and they further decay to produce more stable nuclei. The energy produced in the fission is in the form of kinetic energy of the products, i.e., in the form of heat which can be collected and converted to other forms of energy as needed. Uranium Nuclear Reactor: A nuclear reactor is an apparatus or a device in which nuclear fission is carried out in a controlled manner to produce energy in the form heat which is then converted to electricity. In a uranium reactor, 92 U 235 is used as the fuel. It is bombarded by slow neutrons to produce 92 U 236 which undergoes fission. Example 15.12: Calculate the energy released in the reaction 92 U 236 → I + Y + 2n 53 137 39 97 . The masses of 92 U 236 , 53 I and Y 137 39 97 are 236.04557, 136.91787 and 96.91827 respectively. Solution: Energy released is given by Q = [mU – mI – mY – 2mn ]c2 = [236.04557 – 136.91787 – 96.91827 -2 x 1.00865] c2 = 0.19011 × 931.5 MeV = 177.0875 MeV Chain Reaction: Neutrons are produced in the fission reaction shown in Eq. (15.31). Some reactions produce 2 neutrons while others produce 3 or 4 neutrons. The average number of neutrons per reaction can be shown to be 2.7. These neutrons are in turn absorbed by other 92 U 235 nuclei to produce 92 U 236 which undergo fission and produce further 2.7 neutrons per fission. This can have a cascading effect and the number of neutrons produced and therefore the number of 92 U 236 nuclei produced can increase quickly. This is called a chain reaction. Such a reaction will lead to a fast increase in the number of fissions and thereby in a rapid increase in the amount of energy produced. This will lead to an explosion. In a nuclear reactor, methods are employed to stop a chain reaction from occurring and fission and energy generation is allowed to occur in a controlled fashion. The energy generated, which is in the form of heat, is carried away and converted to electricity by using turbines etc. More than 15 countries have nuclear reactors and use nuclear power. India is one of them. There are 22 nuclear reactors in India,
340 Do you know? the largest one being at Kudankulam, Tamil Nadu. Maximum nuclear power is generated by the USA. 15.11.2. Nuclear Fusion: We have seen that light nuclei (A < 40) have lower EB/A as compared to heavier ones. If any two of the lighter nuclei come sufficiently close, within about one fm of each other, then they can undergo nuclear reaction and form a heavier nucleus. The heavier nucleus will have higher EB/A than the reactants. The mass of the product nucleus will therefore be lower than the total mass of the reactants, and energy of the order of MeV will be released in the process. This process wherein two nuclei fuse together to form a heavier nucleus accompanied by a release of nuclear energy is called nuclear fusion. For a nuclear reaction to take place, it is necessary for two nuclei to come to within about 1 fm of each other so that they can experience the nuclear forces. It is very difficult for two atoms to come that close to each other due to the electrostatic repulsion between the electrons of the two atoms. This problem can be solved by stripping the atoms of their electrons and producing bare nuclei. It is possible to do so by giving the electrons energies larger than the ionization potentials of the atoms by heating a gas of atoms. But even after this, the two bare nuclei find it very difficult to go near each other due to the repulsive force between their positive charges. For nuclear fusion to occur, we have to heat the gas to very high temperature thereby providing the nuclei with very high kinetic energies. These high energies can help them to overcome the electrostatic repulsion and come close to one another. As the positive charge of a nucleus goes on increasing with increase in its atomic number, the kinetic energies of the nuclei, i.e., the temperature of the gas necessary for nuclear fusion to occur goes on increasing with increase in Z. The temperature at the centre of the Sun is about 107 K. The nuclear reactions taking place at the centre of the Sun are the fusion of four hydrogen nuclei, i.e., protons to form a helium nucleus. Of course, because of the electrostatic repulsion and the values of densities at the centre of the Sun, it is extremely unlikely that four protons will come sufficiently close to one another at a given time so that they can combine to form helium. Instead, the fusion proceeds in several steps. The effective reaction can be written as 4 p → α + 2e+ + neutrinos + 26.7 MeV. These reactions have been going on inside the Sun since past 4.5 billion years and are expected to continue for similar time period in the future. At the centres of other stars where temperatures are higher, nuclei heavier than hydrogen can fuse generating energy. Nuclear fusion is taking place all the time in the universe. It mostly takes place at the centres of stars where the temperatures are high enough for nuclear reactions to take place. There, light nuclei fuse into heavier nuclei generating energy in the process. Nuclear fusion is in fact the source of energy for stars. Most of the elements heavier than boron till iron, that we see around us today have been produced through nuclear fusion inside stars. Light elements, i.e., deuterium, helium, lithium, beryllium and boron, have not been created inside stars, but are believed to have been created within the first 200 second in the life of the universe, i.e., within 200 seconds of the big bang which marked the beginning of the universe. The temperature at that time was very high and some nuclear reactions could take place. After about 200 s, the temperature decreased and nuclear reactions were no longer possible.
341 Internet my friend Do you know? Do you know? The discussion on nuclear energy will not be complete without mentioning its harmful effects. If an uncontrolled chain reaction sets up in a nuclear fuel, an extremely large amount of energy can be generated in a very short time. This fact has been used to produce what are called atom bombs or nuclear devices. Either fission alone or both fission and fusion are used in these bombs. The first such devices were made towards the end of the second world war by America. By now, several countries including India have successfully made and Example 15.13: Calculate the energy released in the fusion reaction taking place inside the Sun, 4 p → α + 2e+ + neutrinos, neglecting the energy given to the neutrinos. Mass of alpha particle being 4.001506 u. Solution: The energy released in the process, ignoring the energy taken by the neutrinos is given by Q m fi ff ffi ffl ffl m mff c fl 4 2 2 p e , Q c fi ff ffi fl ffl ffl ff fi ff fi 4 1 00728 4 001506 2 0 00055 0 026514 931 5 2 . . . . . .24 698 MeV 1. https://www.siyavula.com/read/ science/grade-10/the-atom/04-theatom-02 2. https://en.wikipedia.org/wiki/Bohr_ model 3. http://hyperphysics.phy-astr.gsu.edu/ hbase/quantum/atomstructcon.html 4. https://en.wikipedia.org/wiki/Atomic_ nucleus 5. https://en.wikipedia.org/wiki/ Radioactive_decay The fusion inside stars can only take place between nuclei having mass number smaller than that of iron, i.e., 56. The reason for this is that iron has the highest EB/A value among all elements as seen from Fig.15.6. If an iron nucleus fuses into another nucleus, the atomic number of the resulting nucleus will be higher than that of iron and hence it will have smaller EB/A. The mass of the resultant nucleus will hence be larger than the sum of masses of the reactants and energy will have to be supplied to the reactants for the reaction to take place. The elements heavier than iron which are present in the universe are produced via other type of nuclear reaction which take place during stellar explosions. We have seen that the activity of radioactive material decreases exponentially with time. Other examples of exponential decay are • Amplitude of a simple pendulom decays exponentially as A =A0 e-bt, where b is damping factor. • Cooling of an object in an open surrounding is exponential. Temperature θ =θ 0 e-kt where k depends upon the object and the surrounding. • Discharging of a capacitor through a pure resistor is exponential. Charge Q on the capacitor at a given instant is Q = Q0 e-[t/RC] where RC is called time constant. • Charging of a capacitor is also exponential but, it is called exponential growth. tested such nuclear devices. America remains the only country to have actually used two atom bombs which completely destroyed the cities of Hiroshima and Nagasaki in Japan in early August 1945.